\magnification=\magstep1 \newfam\msbfam \font\tenmsb=msbm10 \textfont\msbfam=\tenmsb \font\sevenmsb=msbm7 \scriptfont\msbfam=\sevenmsb \def\Bbb#1{{\fam\msbfam\relax#1}} \def\N{{\Bbb N}} \def\Q{{\Bbb Q}} \def\R{{\Bbb R}} \def\Z{{\Bbb Z}} \def\barf{{\overline{f}}} \def\barK{{\overline{K}}} \def\barS{{\overline{S}}} \def\barU{{\overline{U}}} \def\barV{{\overline{V}}} \def\scra{{\cal A}} \def\scrb{{\cal B}} \def\scrc{{\cal C}} \def\scrd{{\cal D}} \def\scre{{\cal E}} \def\scrf{{\cal F}} \def\scrg{{\cal G}} \def\scri{{\cal I}} \def\scrk{{\cal K}} \def\scrm{{\cal M}} \def\scrn{{\cal N}} \def\scro{{\cal O}} \def\scrp{{\cal P}} \def\scrq{{\cal Q}} \def\scrr{{\cal R}} \def\scrs{{\cal S}} \def\scru{{\cal U}} \def\scrv{{\cal V}} \def\scrw{{\cal W}} \def\scrts{{\cal T\cal S}} \def\scrpts{{\cal P\cal T\cal S}} \def\scrfb{{\cal F\cal B}} \def\scrns{{\cal N\cal S}} \def\scrtns{{\cal T\cal N\cal S}} \def\scrdd{{\cal D\cal D}} \def\hatc{{\widehat{c}}} \def\hatf{{\widehat{f}}} \def\hatg{{\widehat{g}}} \def\hath{{\widehat{h}}} \def\hatE{{\widehat{E}}} \def\hatX{{\widehat{X}}} \def\hatalpha{{\widehat{\alpha}}} \def\hatbeta{{\widehat{\beta}}} \def\hatgamma{{\widehat{\gamma}}} \def\hatalphabeta{{\widehat{\alpha\|\beta}}} \def\tildef{{\widetilde{f}}} \def\tildeg{{\widetilde{g}}} \def\tildeh{{\widetilde{h}}} \def\tildez{{\widetilde{z}}} \def\tildeW{{\widetilde{W}}} \def\tildeX{{\widetilde{X}}} \def\tildeY{{\widetilde{Y}}} \def\tildeZ{{\widetilde{Z}}} \def\Cl{{\rm Cl}} \def\Int{{\rm Int}} \def\Bd{{\rm Bd}} \def\Hom{{\rm Hom}} \def\Arr{{\rm Arr}} \def\Dom{{\rm Dom}} \def\Tar{{\rm Tar}} \def\dom{{\rm dom}} \def\tar{{\rm tar}} \def\Obj{{\rm Obj}} \def\id{{\rm id}} \def\Id{{\rm Id}} \def\catofsets{\{\hbox{Sets}\}} \def\catofrings{\{\hbox{Rings}\}} \def\catofgps{\{\hbox{Groups}\}} \def\set{{{\rm set}}} \def\SL{{{\rm SL}}} \def\Stab{{{\rm Stab}}} \def\det{{{\rm det}}} \def\supp{{{\rm supp}}} \centerline{\bf Notes for Math 8301, Manifolds and Topology, Fall 2004} \centerline{\bf The Fundamental Group and Covering Spaces} \vskip.1in\noindent {\it Definition.} $B_\delta(x)=(x-\delta,x+\delta)\subseteq\R$. \vskip.1in\noindent {\it Definition.} A function $f:\R\to\R$ is {\bf continuous at $x$} if, for all $\epsilon>0$, there exists $\delta>0$ such that $f(B_\delta(x))\subseteq B_\epsilon(f(x))$, {\it i.e.}, $B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))$, \vskip.1in\noindent {\it Note:} For $\phi:A\to B$ and $B_0\subseteq B$, we define $\phi^{-1}(B_0)=\{a\in A\,|\,\phi(a)\in B_0\}$. \vskip.1in\noindent {\it Definition.} A function $f:\R\to\R$ is {\bf continuous} if, for all $x\in\R$, we have: $f$ is continuous at $x$. \vskip.1in\noindent {\it Question:} How do we define continuity for $f:\R^m\to\R^n$? How about $f:S^n\to\R^m$, where $S^n:=\{x\in\R^{n+1}\,|\,d(x,0)=1\}$? How about $f:S^n\to S^m$? Let's try to formulate a general definition of continuity that will include all these definitions as special cases. \vskip.1in\noindent {\it Definition.} An {\bf$N$-space} consists of \itemitem{(1)} a set $X$ and \itemitem{(2)} a function $x\mapsto\scrn_x:X\to \{\hbox{subsets of }\{\hbox{subsets of }X\}\}$ \noindent such that \itemitem{(A)} for all $x\in X$, we have: $\scrn_x\ne\emptyset$; and \itemitem{(B)} for all $x\in X$, for all $N\in\scrn_x$, we have: $x\in N$. \vskip.1in\noindent {\it Example.} $X=\R$, $\scrn_x=\{B_\delta(x)\,|\,\delta>0\}$. \vskip.1in\noindent {\it Definition.} For all $x\in X$, an {\bf ``$N$-neighborhood''} of $x$ is an element of $\scrn_x$. \vskip.1in\noindent {\it Definition.} Let $X$ and $Y$ be $N$-spaces and let $f:X\to Y$ be a function. For a point $x\in X$, we say that $f:X\to Y$ is {\bf $N$-continuous at $x$} if, for any $N$-neighborhood $V$ of $f(x)$, there exists an $N$-neighborhood $U$ of $x$ such that $f(U)\subseteq V$, {\it i.e.}, $U\subseteq f^{-1}(V)$. We say that $f$ is {\bf $N$-continuous} if $f$ is $N$-continuous at every point $x\in X$. \vskip.1in\noindent {\it Definition.} A category consists of \itemitem{(1)} a class $\scrc$ of {\bf objects}; \itemitem{(2)} for all $C,C'\in\scrc$, a set $\Hom(C,C')$ of {\bf morphisms} or {\bf arrows} from $C$ to $C'$; \itemitem{(3)} for all $C,C',C''\in\scrc$, a {\bf composition} function $$\Hom(C,C')\times\Hom(C',C'')\to\Hom(C,C'');\qquad\hbox{and}$$ \itemitem{(4)} for all $C\in\scrc$, an {\bf identity} arrow $\id_C\in\Hom(C,C)$ \noindent such that \itemitem{(A)} for all $C,C'\in\scrc$, for all $f\in\Hom(C,C')$, we have $f\circ\id_C=\id_{C'}\circ f=f$; \itemitem{(B)} for all $C,C',C'',C'''\in\scrc$, for all $f\in\Hom(C,C')$, for all $g\in\Hom(C',C'')$, for all $h\in\Hom(C'',C''')$, we have $(f\circ g)\circ h=f\circ(g\circ h)$; \noindent and probably other properties. \vskip.1in\noindent {\bf EXERCISE 1A:} Look up, and write out the defintion of a category. \vskip.1in\noindent {\it Example.} $\scrns:=\{N\hbox{-spaces}\}$. That is, $\scrns$ is the category of $N$-spaces, together with $N$-continuous maps, together with the usual composition of maps, together with the usual identity maps. \vskip.1in\noindent {\bf EXERCISE 1B:} \itemitem{(1)} For all $X\in\scrns$, show that $x\mapsto x:X\to X$ is $N$-continuous. \itemitem{(2)} For all $X,Y,Z\in\scrns$, for all $N$-continuous $f:X\to Y$, for all $N$-continuous $g:Y\to Z$, show that $g\circ f:X\to Z$ is $N$-continuous. \vskip.1in\noindent Let $\catofsets$ denote the category of sets (and functions). For any $X\in\scrns$, let $X_\set$ be the underlying set of $X$. For $X,Y\in\scrns$, for $f\in\Hom(X,Y)$, let $f_\set:X_\set\to Y_\set$ be the underlying set function of $f$. Then $X\mapsto X_\set:\scrns\to\catofsets$ is an example of a {\bf functor}. (Really, we should say $X\mapsto X_\set$ together with $f\mapsto f_\set$ forms a functor.) Note that, for $X,Y,Z\in\scrns$, for $f\in\Hom(X,Y)$, for $g\in\Hom(Y,Z)$, we have $(g\circ f)_\set=g_\set\,\circ\,f_\set$. This is one of the basic properties of a functor. \vskip.1in\noindent {\bf EXERCISE 1C:} Look up, and write down the definition of a functor from one category to another. Be sure to note that some functors are covariant, while others are contravariant. \vskip.1in\noindent {\it Example.} Let $\catofrings$ denote the category of (not necessarily commutative) rings with unit, together with unit-preserving ring homomorphisms. Let $\catofgps$ denote the category of groups, with group homomorphisms. Then $(R,\cdot,+)\mapsto(R,+)$ is a functor from $\catofrings$ to $\catofgps$ (and even takes values in the subcategory of Abelian groups). Then $(R,\cdot,+)\mapsto(\{\hbox{units in }R\},\cdot)$ is another functor from $\catofrings$ to $\catofgps$. \vskip.1in\noindent {\it Definition.} Let $\scrc$ be any category. Then there is an identity functor $\Id_\scrc:\scrc\to\scrc$ defined by $\Id_\scrc(C)=C$. (Note: This defines the functor on objects, and it is left to you you to guess what the functor does on arrows.) (Note: Well, okay, I'll tell you this time -- it leaves each arrow fixed, {\it i.e.}, for all $f:C\to C'$ in $\scrc$, we define $\Id_\scrc(f)=f$. In general, though, it is an expectation that you should be able to figure this kind of thing out, but always feel free to ask if it's unclear.) \vskip.1in\noindent Fix, until END OF DISCUSSION \#1, some $X\in\scrns$. Let's say we've identified some points of $X$ that are ``happy''. \vskip.1in\noindent Let $0\in X$. If we say {\bf ``points close to $x$ are happy''} or {\bf ``points sufficiently close to $x$ are happy''} or {\bf ``all points sufficiently close to $x$ are happy''}, we mean that there is an $N$-neighborhood $U$ of $x$ such that, for all $u\in U$, we have: $u$ is happy. Now suppose that $y$ is sufficiently close to $x$ that it is guaranteed to be happy. That is, suppose that $y\in U$. {\it Question:} Does it automatically follow that all points sufficiently close to $y$ are guaranteed to be happy by being in $U$? In other words, can we conclude that there is a neighborhoos $V$ of $y$ such that $V\subseteq U$. \vskip.1in\noindent {\it Answer:} No. For example, let $X_\set=\R$, and for all $x\in X$, let $\scrn_x:=\{[x-\delta,x+\delta]\,|\,\delta>0\}$. Then $X=(X_\set,x\mapsto\scrn_x)$ is an $N$-space. Suppose that $[-1,1]$ is exactly the set of happy points. Let $x:=0, y:=-1$. Then $U=[-1,1]$ is an $N$-neighborhood of $x$ and $y\in U$, but there is no $N$-neighborhood $V$ of $y$ such that $V\subseteq U$. So even though $y$ is happy, we have no ``room to maneuver'' because the kinds of $N$-neigborhoods in $X$ are, somehow, very {\it unforgiving}, in the sense that, if you have a point that is in an $N$-neighborhood, and you then perturb it ever so slightly, you may find yourself outside the $N$-neighborhood. \vskip.1in\noindent {\it Definition.} We say that an $N$-space $X$ is {\bf forgiving} if, for all $x\in X$, for any $N$-neighborhood $U$ of $x$, for any $y\in U$, there is an $N$-neighborhood $V$ of $y$ such that $V\subseteq U$. \vskip.1in\noindent END OF DISCUSSION \#1 \vskip.1in\noindent Fix, until END OF DISCUSSION \#2, some $X\in\scrns$. Let's say we've identified some points of $X$ that are ``happy'' and some point that are ``dull''. So $X$ breaks up into happy dull points, unhappy dull points, happy sharp points and unhappy sharp points. Let $x\in X$ and assume that all points sufficiently close to $x$ are happy. That is, there is an $N$-neighborhood $U$ of $x$ such that every element of $U$ is happy. Assume, furthermore that all points sufficiently close to $x$ are dull. That is, there is an $N$-neighborhood $V$ of $x$ such that every element of $V$ is happy. Does it automatically follow that there is a neighborhood $W$ of $x$ such that $W\subseteq U\cup V$, which guarantees that all points sufficiently close to $x$ are both happy and dull? The answer is no. It's not an assigned problem, but you should think about constructing a counterexample. \vskip.1in\noindent {\it Definition.} An $N$-space $X$ is {\bf intersection-stable} if, for all $x\in X$, for any $N$-neighborhoods $U$ and $V$ of $x$, there is an $N$-neighborhood $W$ of $x$ such that $W\subseteq U\cap V$. \vskip.1in\noindent END OF DISCUSSION \#2 \vskip.1in\noindent {\it Definition.} An $N$-space is said to be {\bf topological} if it is both forgiving and intersection-stable. \vskip.1in\noindent {\it Definition.} Let $\scrtns$ denote the category of topological $N$-spaces (with $N$-continuous maps). \vskip.1in\noindent {\it Definition.} Let $X$ be a set. A {\bf topology on $X$} is a subset $\scro$ of $\{\hbox{subsets of }X\}$ such that \itemitem{(A)} $\emptyset,X\in\scro$; \itemitem{(B)} for all $\scrs\subseteq\scro$, we have $\cup\scrs\in\scro$; and \itemitem{(C)} for all finite $\scrf\subseteq\scro$, we have $\cap\scrf\in\scro$. \vskip.1in\noindent {\it Definition.} A {\bf topological space} consists of \itemitem{(1)} a set $X$; and \itemitem{(2)} a topology on $X$. \vskip.1in\noindent {\it Definition.} An {\bf open subset} of $X$ is an element of $\scro$. A subset of $X$ is said to be {\bf open} or {\bf open in $X$} if it is an open subset of $X$. \vskip.1in\noindent {\it Definition.} Let $X$ and $Y$ be topological spaces and let $f:X\to Y$ be a function defined on their underlying sets. We say that $f$ is {\bf continuous} if, for any open $V\subseteq Y$, we have that $f^{-1}(V)$ is open in $X$. \vskip.1in\noindent What motivates the definition of topological space and of this definition of continuity? \vskip.1in\noindent {\it Definition.} Let $\scrts$ denote the category of topological spaces, with continuous maps. \vskip.1in\noindent {\it Definition.} Define a functor $\scrf:\scrtns\to\scrts$ by $$\scrf((X,x\mapsto\scrn_x))= (X,\{\cup\scrs\,|\,\scrs\subseteq\cup_{x\in X}\scrn_x\}).$$ Here, I leave it to you to guess what $\scrf$ does on arrows. Note that $\{\hbox{open sets in }\scrf X\}$ is exactly the closure of $\{N\hbox{-neighborhoods of points of }X\}$ under union. Define a functor $\scrg:\scrts\to\scrtns$ by $$\scrg((Y,\scro))= (Y,y\mapsto\{U\in\scro\,|\,y\in U\}).$$ \vskip.1in\noindent {\bf EXERCISE 1D:} Show, for all $X\in\scrtns$, that $\scrf X\in\scrts$. (You must show that the open subsets of $\scrf X$ are closed under finite intersection and arbitrary union, and that both $\emptyset$ and $X$ are open in $\scrf X$.) \vskip.1in\noindent {\bf EXERCISE 1E:} Show, for all $Y\in\scrts$, that $\scrg Y\in\scrtns$. \vskip.1in\noindent {\bf EXERCISE 1F:} Show, for all $Y\in\scrts$, that $\scrf\scrg Y=Y$. (You must show that a subset of $Y$ is open in $\scrf\scrg Y$ iff it is open in $Y$.) \vskip.1in\noindent The preceding exercise asks you to prove that $\scrf\scrg=\Id_\scrts$. We now show that $\scrg\scrf X$ is not necessarily equal to $X$. That is, we show that $\scrf\scrg\ne\Id_\scrtns$: \vskip.1in\noindent {\it Example.} Let $X:=(\R,x\mapsto\{(x-\delta,x+\delta)\,|\,\delta>0\})$. Then the set of open subsets of $\scrf X$ is exactly the closure of $\{(x-\delta,x+\delta)\,|\,x\in\R,\delta>0\}$ under union. In particular, $(-1,1)\cup(2,4)$ is open in $\scrf X$. Then $(-1,1)\cup(2,4)$ is an $N$-neighborhood of $0$ in $\scrg\scrf X$. However it is {\it not} an $N$-neighborhood of $0$ in $X$. Therefore $\scrg\scrf X\ne X$. \vskip.1in\noindent {\it Definition.} An arrow (or morphism) $f:C\to C'$ in a category $\scrc$ is an {\bf isomorphism} if there exists an arrow $g:C'\to C$ such that $f\circ g=\id_{C'}$ and such that $g\circ f=\id_C$. \vskip.1in That is, an isomorphism is simply an invertible arrow. \vskip.1in\noindent {\it Definition.} In the category $\scrts$, an isomorphism will be called a {\bf homeomorphism}. In the category $\scrtns$, an isomorphism will be called an {\bf$N$-homeomorphism}. \vskip.1in\noindent {\it Definition.} Let $\scra$ and $\scrb$ be functors from a category $\scrc$ to a category $\scrd$. A {\bf natural transformation} $\tau:\scra\to\scrb$ associates to each object $C\in \scrc$ an arrow $\tau_C:\scra C\to\scrb C$, provided that this association has the property that: for any arrow $f:C\to C'$ in $\scrc$, we have $(\scrb f)\circ\tau_C=\tau_{C'}\circ(\scra f)$. \vskip.1in Note that, for their to be a natural transformation from $\scra$ to $\scrb$, the domains of $\scra$ and $\scrb$ must agree. Moreover, their targets must agree as well. Let $\scra$ and $\scrb$ be functors from a category $\scrc$ to a category $\scrd$. It may be helpful to picture $\scra\scrc$ as a subcategory inside of $\scrd$. In some sense, $\scra$ is a ``parametric'' subcategory of $\scrd$, where the objects are parametrized by objects in $\scrc$. Similarly, $\scrb\scrc$ is a parametric subcategory of $\scrd$. With this intution, a transformation $\tau:\scra\to\scrb$ may be thought of as a parametric family of arrows in $\scrd$, one for each object in $\scrc$. For each $C\in\scrc$, the corresponding arrow runs from $\scra C$ to $\scrb C$. Moreover, for each arrow in $\scrc$, we now get a diagram in $\scrd$ (with four objects and four arrows). If these diagrams all commute, then we say that the transformation is ``natural''. \vskip.1in\noindent {\it Example.} Let $\scrv$ be the category of finite dimensional vector spaces, and linear transformations. Let $\scri:\scrv\to\scrv$ be the (covariant) identity functor. Let $\scrd:\scrv\to\scrv$ be the (contravariant) functor defined by $\scrd(V)=V^*:=\Hom(V,\R)$. Let $\scrdd:\scrv\to\scrv$ be the (covariant) functor $\scrdd:=\scrd\circ\scrd$ obtained by composing $\scrd$ with itself. Then, for all $V\in\scrv$, we have $\scrdd(V)=V^{**}$. For all $V\in\scrv$, let $\tau_V:V\to V^{**}$ be defined by $(\tau_Vv)(l)=l(v)$. (Note: The map $\tau_Vv:V^*\to\R$ is sometimes called the ``evaluational at $v$''.) Then $\tau:\scri\to\scrdd$ is a natural tranformation, as Exercise 2B will verify. By contrast, the next exercise shows that there is no natural transformation from $\scri$ to $\scrd$, even though, for all $V\in\scrv$, the vector spaces $V$ and $V^*$ are isomorphic, since they have the same dimension. \vskip.1in\noindent {\bf EXERCISE 2A:} Let $V:=\R^2$ and let $f:V\to V^*$ be an isomorphism. For any linear map $g:V\to V$, let $g^*:V^*\to V^*$ be defined by $g^*(l)=l\circ g$. Show that there exists an isomorphism $g:V\to V$ such that $f\ne g^*\circ f\circ g$. \vskip.1in\noindent {\bf EXERCISE 2B:} As in the example above, for any real vector space $V$, let $\tau_V:V\to V^{**}$ be defined by $(\tau_Vv)(l)=l(v)$. Let $W$ and $X$ be real vector spaces and let $g:W\to X$ be a linear transformation. Define $g^*:X^*\to W^*$ by $g^*(l)=l\circ g$ and define $g^{**}:W^{**}\to X^{**}$ by $g^{**}(l)=l\circ g^*$. Show that $g^{**}\circ\tau_W=\tau_X\circ g$. \vskip.1in\noindent {\it Definition.} Let $\scrc$ be a category. Then the {\bf arrow category of $\scrc$}, $\Arr(\scrc)$, is a category whose objects are the arrows in $\scrc$, and for which, for all pairs of arrows $f:C_1\to C_2$, $f':C'_1\to C'_2$ in $\scrc$, we define $\Hom(f,f')$ to be the collection of all those $$(g_1,g_2)\in\Hom(C_1,C'_1)\times\Hom(C_2,\times C'_2)$$ such that $g_2\circ f=f'\circ g_1$. \vskip.1in We leave it as an exercise to guess what the definitions of compositions and identity arrows in $\Arr(\scrc)$. \vskip.1in\noindent {\it Definition.} Given a category $\scrc$, we define two functors $\Dom_\scrc,\Tar_\scrc:\Arr(\scrc)\to\scrc$ by: for all $f:D\to T$ in $\scrc$, we set $\Dom_\scrc(f)=D$ and $\Tar_\scrc(f)=T$. \vskip.1in With this terminology in place, if $\scra,\scrb:\scrc\to\scrd$ are two functors, then a natural transformation $\mu:\scra\to\scrb$ is equivalent to a functor $\scrm:\scrc\to\Arr(\scrd)$ satisfying both $\Dom_\scrd\circ\scrm=\scra$ and $\Tar_\scrd\circ\scrm=\scrb$. \vskip.1in Now recall the functors $\scrf:\scrtns\to\scrts$ and $\scrg:\scrts\to\scrtns$ defined above. Recall that $\scrf\scrg$ is the identity functor on $\scrts$, but that $\scrg\scrf$ is not the identity functor on $\scrtns$. We will make clear next a sense in which $\scrf\scrg$ is ``close'' to the identity on $\scrtns$. \vskip.1in\noindent {\it Definition.} Let $\scrc$ and $\scrd$ be categories and let $\scra:\scrc\to\scrd$ be a functor. Define a natural transformation $\iota^\scra:\scra\to\scra$ by $\iota^\scra_C:=\id_{\scra C}:\scra C\to\scra C$. This natural transformation is called {\bf the identity on $\scra$}. \vskip.1in\noindent {\it Definition.} Let $\scrc$ and $\scrd$ be categories, let $\scrp,\scrq,\scrr:\scrc\to\scrd$ be functors and let $\alpha:\scrp\to\scrq$ and $\beta:\scrq\to\scrr$ be natural transformations. We define a natural transformation $\beta\alpha:\scrp\to\scrr$ by $(\beta\alpha)_C=\beta_C\circ\alpha_C$. \vskip.1in\noindent {\it Definition.} We say that two functors $\scra,\scrb:\scrc\to\scrd$ are {\bf equivalent} (and write $\scra\sim\scrb$) if there exist natural transformations $\tau:\scra\to\scrb$ and $\mu:\scrb\to\scra$ such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$. \vskip.1in Intuitively, if two functors are equivalent, then they have the same ``information'' in them. If I'm an expert on one of the functors and you're an expert on the other, then we'll probably spend our days quibbling over notation, but will be well aware that we both know the same stuff. \vskip.1in\noindent {\bf EXERCISE 2C:} Let $\scra,\scrb:\scrc\to\scrd$ be functors. Let $\tau:\scra\to\scrb$ be a natural transformation. Suppose, for all $C\in\scrc$, that $\tau_C:\scra C\to\scrb C$ is an isomorphism. Show that there is a unique natural transformation $\mu:\scrb\to\scra$ such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$. \vskip.1in\noindent {\bf EXERCISE 2D:} Let $\scra,\scrb:\scrc\to\scrd$ be functors. Let $\tau:\scra\to\scrb$ and $\mu:\scrb\to\scra$ be natural transformations such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$. Show, for all $C\in\scrc$, that $\tau^C:\scra C\to\scrb C$ is an isomorphism. \vskip.1in\noindent {\bf EXERCISE 2E:} Recall the functors $\scrf:\scrtns\to\scrts$ and $\scrg:\scrts\to\scrtns$ defined above. Recall that $\scrf\scrg=\Id_\scrts$ but that $\scrg\scrf\ne\Id_\scrtns$. Prove that $\scrg\scrf\sim\Id_\scrtns$. \vskip.1in\noindent {\it Definition.} A category $\scrc$ is said to be {\bf equivalent} to a category $\scrd$ (write $\scrc\sim\scrd$) if there are functors $\scrp:\scrc\to\scrd$ and $\scrq:\scrd\to\scrc$ such that $\scrp\scrq\sim\Id_\scrd$ and $\scrq\scrp\sim\Id_\scrc$. \vskip.1in Intuitively, if two categories are equivalent, then they have the same ``information'' in them. If I'm an expert on one of the categories and you're an expert on the other, then we'll probably spend our days quibbling over notation, but will be well aware that we both know the same stuff. The preceding exercise asks you to demonstrate that $\scrtns$ is equivalent to $\scrts$. Which is better, $\scrtns$ or $\scrts$? In some sense, neither, since they're equivalent. On the other hand, with a little thought, you will readily see that $\scrf:\scrtns\to\scrts$ is not ``one-to-one''. To almost any topological space Y, there are several topological $N$-spaces whose images under $\scrf$ are all equal to $Y$. For example, $(\R,x\mapsto\{(x-\delta,x+\delta)\})$ and $(\R,x\mapsto\{(x-2\delta,x+\delta)\})$ have the same image under $\scrf$. So, in some sense, $\scrtns$ is redundant, with more copies of each space than one really needs. From this perspective, $\scrts$ is better than $\scrtns$. Finally, a few broad words about mathematics. Typically a working mathematician develops expertise in a category, {\it e.g.}, the category $\scrts$ of topological spaces. A fundamental question in topology is to determine, given two topological spaces, whether they are homeomorphic, but, more generally, whatever your ``category of choice'' may be, you'll want to be able to form many examples of objects in it, and, in many cases to show that two such examples are not isomorphic in the category. If one can form a functor $\scrf$ that goes from your category of choice $\scrc$ to some other category $\scrd$, and if tools are available for distinguishing objects in $\scrd$, then, given $C,C'\in\scrc$, if you are able to show that $\scrf C$ and $\scrf C'$ are not isomorphic in $\scrd$, then you can immediately conclude that $C$ and $C'$ are not isomorphic in $\scrc$. (See the next exercise.) For example, there is a functor called ``first homology'', and denoted $H_1$ which runs from topological spaces to Abelian groups. Given two topological spaces, if one is able to show that their first homology groups are not isomorphic, then the spaces are not homeomorphic. This kind of argument will be central to much of this course! \vskip.1in\noindent {\bf EXERCISE 2F:} Show that functors carry isomorphisms to isomorphisms. That is, show that, if $\scrf:\scrc\to\scrd$ is a functor, if $f:C\to C'$ is an isomorphism in $\scrc$, then $\scrf f:\scrf C\to\scrf C'$ is an isomorphism in $\scrd$. In the next definition, we make the convention that $\cup\emptyset=\emptyset$. \vskip.1in\noindent {\it Definition.} Let $X$ be a set and let $\scrs\subseteq \{\hbox{subsets of }X\}$. The {\bf union closure of $\scrs$} is $\{\cup\scra\,|\,\scra\subseteq\scrs\}$. The {\bf finite intersection closure in $X$ of $\scrs$} is $\{X\}\cup\{\cap\scra\,|\,\scra\subseteq\scrs, 0<|\scra|<\infty\}$. \vskip.1in\noindent {\it Definition.} Let $X$ be a set and let $\scro$ be a topology on $X$. Let $\scrb\subseteq \{\hbox{subsets of }X\}$. We say that $\scrb$ is a {\bf basis} for $\scro$ if the union closure of $\scrb$ is $\scro$. We say that $\scrb$ is a {\bf subbasis} for $\scro$ if the finite intersection closure in $X$ of $\scrb$ is a basis for $\scro$. \vskip.1in Note that any basis for a topology is a subbasis. \vskip.1in\noindent {\bf EXERCISE 2G:} Let $X$ be a set and let $\scrs\subseteq \{\hbox{subsets of }X\}$. Show that $\scrs$ is a subbasis of a topology on $X$. (That is, show that the union closure of the finite intersection closure in $X$ of $\scrs$ is a topology on $X$.) \vskip.1in\noindent {\bf EXERCISE 2H:} Let $X$ be a set and let $\scrb\subseteq \{\hbox{subsets of }X\}$. Show that $\scrb$ is a basis of a topology on $X$ if and only if, both of the following conditions hold: \itemitem{(1)} $\cup\scrb=X$; and \itemitem{(2)} for all $B,B'\in\scrb$, for all $x\in B\cap B'$, there exists $B''\in\scrb$ such that $x\in B''\subseteq B\cap B'$. \vskip.1in\noindent {\it Definition.} Let $\scro$ and $\scro'$ be topologies on a set $X$. We say that $\scro$ is {\bf finer} (or {\bf stronger}) than $\scro'$ if $\scro\supseteq\scro'$. We say that $\scro$ is {\bf coarser} (or {\bf weaker}) than $\scro'$ if $\scro\subseteq\scro'$. \vskip.1in\noindent {\it Remark.} Let $X$ and $I$ be sets and, for all $i\in I$, let $\scro_i$ be a topology on $X$. Then $\displaystyle{{\mathop\bigcap_{i\in I}}\,\scro_i}$ is a topology on $X$. \vskip.1in The following is a restatement of the preceding remark. \vskip.1in\noindent {\it Remark.} Let $\Xi\subseteq \{\hbox{topologies on }X\}$. Then $\cap\Xi$ is a topology on $X$. The preceding two remarks are asserting that, given a collection of topologies on a set, there exists a unique topology which is finest among: \centerline{those topologies coarser than every topology in the collection.} \vskip.1in\noindent {\it Question:} How about the other way around? That is, given a collection of topologies on a set, does there there exist a unique topology which is coarsest among: \centerline{those topologies finer than every topology in the collection.} \vskip.1in\noindent {\it Answer:} We decided in class that the answer is yes. This can probably be argued using Zorn's lemma, but more directly, one can simply take the union of the topologies in the collection and then take the intersection of all the topologies containing that union. By the preceding two remarks, that intersection is again a topology. Yet another way to say the same thing is that you take the union of the topologies in the collection, then take the finite intersection closure of that union, then take the union closure of {\it that}. \vskip.1in {\it Definition.} Let $X$ be a set and let $$\scrs\subseteq \{\hbox{subsets of }X\}.$$ Then the {\bf topology generated by $\scrs$} is the coarsest topology among those topologies containing $\scrs$. Note that, if $\scrb$ is a basis or subbasis of a topology $\scro$ on a set $X$, then $\scro$ is the topology generated by $\scrb$. \vskip.1in\noindent {\it Definition.} Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function and let $$\scrs\subseteq \{\hbox{subsets of }Y\}.$$ Then we define $f^*\scrs:=\{f^{-1}(S)\,|\,S\in\scrs\}$. \vskip.1in\noindent {\bf EXERCISE 2I:} Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function and let $\scro$ be a topology on $Y$. Show that $f^*\scro$ is a topology on $X$. \vskip.1in\noindent {\it Remark.} Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function and let $\scro$ be a topology on $Y$. Then $f^*\scro$ is the coarsest topology making $f$ continuous. \vskip.1in\noindent {\it Definition.} Let $Y$ be a topological space and let $X\subseteq Y$. Then the {\bf inherited topology on $X$} (or {\bf relative topology on $X$} or {\bf subspace topology on $X$}) is: $$\{X\cap U\,|\,U\hbox{ is open in }Y\}.$$ \vskip.1in\noindent {\it Remark.} Let $Y$ be a topological space and let $X\subseteq Y$. Let $i:X\to Y$ be the inclusion, defined by $i(x)=x$. Let $\scro:=\{\hbox{open subsets of }Y\}$. Then $i^*\scro$ is the inherited topology on $X$. It is the coarsest topology making $i$ continuous. \vskip.1in\noindent {\it Definition.} Let $I$ be a set and, for all $i\in I$, let $X_i$ be a topological space. Let $X:=\displaystyle{{\mathop\prod_{i\in I}}\,X_i}$. For all $i\in I$, let $p_i:X\to X_i$ be the $i$th coordinate projection map. For all $i\in I$, let $\scro_i$ be the set of all open subsets of $X_i$. Then the {\bf product topology} on $X$ is the topology generated by $\displaystyle{{\mathop\bigcup_{i\in I}}\,p_i^*\scro_i}$. \vskip.1in Note that the product topology is the coarsest among those that make all the projection maps $p_i$ continuous. In general, whenever a product of topological spaces is formed, we will give it the product topology, unless otherwise specified. In particular, when $X$ and $Y$ are topological spaces, $X\times Y$ is a topological space as well. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space, let $Y$ be a set and let $f:X\to Y$ be a function. The {\bf quotient topology} on $Y$ is the finest making $f$ continuous. (See Exercise 3A below.) \vskip.1in Now let's start making some topological spaces! \vskip.1in\noindent {\it Definition.} Let $n\ge0$ be an integer. The {\bf standard topology} on $\R^n$ is the topology generated by $\{B_r(x)\,|\,x\in\R^n,r>0\}$. \vskip.1in\noindent {\it Definition.} Let $n\ge0$ be an integer. We define $S^n:=\{x\in\R^{n+1}\,|\,d(x,0)=1\}$. The {\bf standard topology} on $S^n$ is the topology inherited from $\R^{n+1}$. \vskip.1in\noindent {\it Definition.} If $X$ is a topological space, and if $x\in X$, then an {\bf open neighborhood of $x$ in $X$} is an open subset $U$ of $X$ such that $x\in U$. If $X$ is a topological space, and if $S\subseteq X$, then an {\bf open neighborhood of $S$ in $X$} is an open subset $U$ of $X$ such that $S\subseteq U$. \vskip.1in\noindent {\it Definition.} To say that a topological space $X$ is {\bf Hausdorff} means: for all $x,x'\in X$, if $x\ne x'$, then there are open neighborhoods $U$ of $x$ and $U'$ of $x'$ in $X$ such that $U\cap U'=\emptyset$. \vskip.1in\noindent {\it Definition.} Let $S$ be a subset of a topological space $X$. We say that $S$ is {\bf closed in $X$} if $X\backslash S$ is open in $X$. \vskip.1in Note that $\{\hbox{closed subsets of }X\}$ is closed under finite union and arbitrary intersection. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space and assume, for all $x\in X$, that $\{x\}$ is closed in $X$. We say that $X$ is {\bf regular} if the following condition holds: For all $x_0\in X$, for any closed subset $C_0$ of $X$, if $x_0\ne C_0$, then there exist open nieghborhoods $U$ and $V$ of $x_0$ and $C_0$ in $X$ such that $U\cap V=\emptyset$. We say that $X$ is {\bf normal} if the following condition holds: For any closed subsets $C,C'$ of $X$, if $C\cap C'=\emptyset$, then there exist open nieghborhoods $U$ and $U'$ of $C$ and $C'$ in $X$ such that $U\cap U'=\emptyset$. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space and let $S\subseteq X$. The {\bf closure} of $S$ in $X$ is the intersection of all closed subsets of $X$ that contain $S$, and is denoted $\barS$ or $\Cl_X(S)$. The {\bf interior} of $S$ in $X$ is the union of all open subsets of $X$ contained in $S$, and is denoted $S^\circ$ or $\Int_X(S)$. The {\bf boundary} of $S$ in $X$ is $\barS\backslash S^\circ$, and is denoted $\partial S$ or $\Bd_X(S)$. \vskip.1in Since an intersection of closed sets is closed and a union of open sets is open, it follows that the closure of $S$ is the smallest closed set containing $S$, while the interior of $S$ is the largest open set contained in $S$. Let $S$ be a subset of a topological space $X$. We say that $S$ is {\bf dense} in $X$ if $\Cl_X(S)=X$. \vskip.1in\noindent {\it Definition.} We say that a topological space is {\bf discrete} if all of its subsets are open. We say that a topological space $X$ is {\bf indiscrete} if its only open sets are $\emptyset$ and $X$. \vskip.1in\noindent {\it Definition.} Let $X$ be a Hausdorff topological space. We say $X$ is {\bf compact} if, for any $\scrs\subseteq\{\hbox{open subsets of }X\}$, we have: \itemitem{($*$)} if $\cup\scrs=X$, then there is a finite subset $\scrf$ of $\scrs$ such that $\cup\scrf=X$. \vskip.1in That is, a Hausdorff topological space is said to be compact if every open cover has a finite subcover, in the following terminology: \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space. An {\bf open cover} of $X$ is a subset $\scru$ of $\{\hbox{open subsets of }X\}$ such that $\cup\scru=X$. Given two covers $\scru,\scrv$ of $X$, we say that $\scru$ is a {\bf subcover} of $\scrv$ if $\scru\subseteq\scrv$. \vskip.1in\noindent {\it Question:} Is $S^2$ homeomorphic to $\R^2$? \vskip.1in\noindent {\it Fact.} $S^2$ is compact, whereas $\R^2$ is not. \vskip.1in\noindent {\it Answer to preceding question:} No. \vskip.1in\noindent {\it Definition.} Let $X$ be a Hausdorff topological space. Then $X$ is said to be {\bf locally compact} if, for all $x\in X$, there exists an open neighborhood $U$ of $x$ in $X$ such that $\barU$ is compact. \vskip.1in\noindent {\it Example.} $\R$ is locally compact, and any finite product of locally compact topological spaces is again locally compact. Consequently, for all integers $n\ge1$, $\R^n$ is locally compact. However infinite products of locally compact topological spaces are generally {\it not} locally compact, {\it e.g.}, $\displaystyle{{\mathop\prod_{i=1}^\infty}\,\R}$ is not locally compact. \vskip.1in\noindent {\it Next Question:} Is $\R$ homeomorphic to $\R^2$? \vskip.1in\noindent {\bf EXERCISE 2J:} For all $p,q\in\R^2$, show that $\R^2\backslash\{p\}$ is homeomorphic to $\R^2\backslash\{q\}$. \vskip.1in\noindent {\bf EXERCISE 2K:} Let $X$ and $Y$ be topological spaces and let $x\in X$. Assume that $X$ is homeomorphic to $Y$. Show that there exists $y\in Y$ such that $X\backslash\{x\}$ is homeomorphic to $Y\backslash\{y\}$. \vskip.1in\noindent {\it Definition.} Let $S$ be a subset of a topological space $X$. We say that $S$ is {\bf clopen in $X$} if $S$ is both closed an open in $X$. \vskip.1in\noindent {\it Definition.} We say that a topological space $X$ is {\bf connected} if it has no clopen sets other than $\emptyset$ and $X$. A topological space is {\bf disconnected} if it is not connected. \vskip.1in\noindent {\it Fact.} $\R\backslash\{0\}$ is disconnected, while $\R^2\backslash\{(0,0)\}$ is connected. \vskip.1in\noindent {\it Remark.} $\R$ is not homeomorphic to $\R^2$. \vskip.1in\noindent {\it Proof:} Assume, for a contradiction that $\R$ and $\R^2$ are homeomorphic. By Exercise 2K, choose $y\in\R^2\backslash\{(0,0)\}$ such that $\R\backslash\{0\}$ is homeomorphic to $\R^2\backslash\{y\}$. By Exercise 2J, $\R^2\backslash\{y\}$ is homeomorphic to $\R^2\backslash\{(0,0)\}$. Then $\R\backslash\{0\}$ is homeomorphic to $\R^2\backslash\{(0,0)\}$. However, by the fact above, $\R\backslash\{(0\}$ is disconnected, while $\R^2\backslash\{(0,0)\}$ is connected. {\bf QED} \vskip.1in In the above proof, we used, without comment, the fact that if two topological spaces are homeomorphic and if one is connected, than the other is as well. That is, ``connectedness is a homeomorphism invariant''. In fact, for a property of topological spaces to be useful, it is typically a homeomorphism invariant, and we will leave it as an implicit exercise to verify that the properties we may introduce below are, in fact, homeomorphism invariants. Typically, such arguments are straightforward. \vskip.1in\noindent {\it Next Question:} Is $\R^2$ homeomorphic to $\R^3$? \vskip.1in Note that both $\R^2\backslash\{(0,0)\}$ and $\R^3\backslash\{(0,0,0)\}$ are connected. Nevertheless, some variant of the argument given above will work to show that the answer to the above question is ``no''. \vskip.1in\noindent {\it Definition.} A {\bf pointed topological space} consists of \itemitem{(1)} a topological space $X$; and \itemitem{(2)} a point $x\in X$. \vskip.1in\noindent We call $x$ the {\bf basepoint} of the pointed topological space $(X,x)$. \vskip.1in\noindent {\it Definition.} The category of pointed topological spaces (with basepoint preserving continuous maps) is denoted $\scrpts$. The forgetful functor $(X,x)\mapsto X\scrpts\to\scrts$ is denoted by $\scrfb$. An isomorphism in $\scrpts$ is called a {\bf basepoint-preserving homeomorphism}. \vskip.1in The next Fact is nontrivial, but we will assume it for now, and prove it in the next few lectures. \vskip.1in\noindent {\it Fact.} There is a functor $\pi_1:\scrpts\to\{\hbox{groups}\}$ such that: \itemitem{(1)} for all $x\in X:=\R^2\backslash\{(0,0)\}$, we have that $\pi_1(X,x)$ is isomorphic to the additive group of integers; and \itemitem{(2)} for all $y\in Y:=\R^3\backslash\{(0,0,0)\}$, we have that $\pi_1(Y,y)$ is isomorphic to the trivial group. \vskip.1in\noindent {\it Remark.} For all topological spaces $X,Y$, for all $x\in X$, if $X$ is homeomorphic to $Y$ then there exists $y\in Y$ such that $(X,x)$ is basepoint-preserving homeomorphic to $(Y,y)$. \vskip.1in\noindent {\it Remark.} Let $X:=\R^2\backslash\{(0,0)\}$ and $Y:=\R^3\backslash\{(0,0,0)\}$. Then $X$ and $Y$ are not homeomorphic. \vskip.1in\noindent {\it Proof:} Say for a contradiction that $X$ is homeomorphic to $Y$. Choose $x\in X$. By the second to last Remark, choose $y\in Y$ such that $(X,x)$ is basepoint-preserving homeomorphic to $(Y,y)$. Let $\pi_1:\scrpts\to\{\hbox{groups}\}$ be the functor described in the preceding Fact. Then, by Exercise 2F, $\pi_1(X,x)$ is isomorpic to $\pi_1(Y,y)$, in the category of groups. We say that $\pi_1(X,x)$ is the {\bf fundamental group of $X$} (with respect to $x$). Then, by the preceding fact, the additive group of integers is isomorphic to the trivial group, contradiction. {\bf QED} \vskip.1in\noindent {\it Corollary.} $\R^2$ is not homeomorphic to $\R^3$. \vskip.1in Given the preceding Remark, the preceding Corollary is argued exactly as in the proof that $\R$ and $\R^2$ are not homeomorphic. Our goal now becomes to define and study the functor $\pi_1:\scrpts\to\{\hbox{groups}\}$ of the preceding Fact. \vskip.1in\noindent {\it Definition.} A space $X$ is {\bf locally compact} if every point of $X$ has an open neighborhood whose closure in $X$ is compact. \vskip.1in\noindent {\it Definition.} Let $X$ and $Y$ be topological spaces and assume that $X$ locally compact. Recall that $\Hom(X,Y)$ is the set of continuous maps from $X$ to $Y$. Let $\scrk$ denote the set of all compact subsets of $X$. Let $\scro$ denote the set of all open subsets of $Y$. For all $K\in\scrk$, for all $U\in\scro$, let let $\scrw(K,U):=\{f\in\Hom(X,Y)\,|\,f(K)\subseteq U\}$. The topology on $\Hom(X,Y)$ generated by the subbasis $\{\scrw(K,U)\,|\,K\in\scrk,U\in\scro\}$ is called the {\bf compact-open} topology on $\Hom(X,Y)$. We let $C(X,Y)$ denote the topological space whose underlying set is $\Hom(X,Y)$ and whose topology is the compact-open topology. \vskip.1in One has the general feeling that a space is locally compact if it is ``almost'' finite dimensional. In particular, if $I$ is a set and, for all $i\in I$, we have a Hausdorff topological space $X_i$, then $\displaystyle{{\mathop\prod_{i\in I}}\,X_i}$ is locally compact iff $\{i\in I\,|\,X_i\hbox{ is noncompact}\}$ is finite. The topological space $C(\R,\R)$ does not feel anywhere near finite dimensional, so one would guess that it is not locally compact, and that is, in fact true: \vskip.1in\noindent {\it Remark.} $C(\R,\R)$ is not locally compact. \vskip.1in\noindent {\it Sketch of proof:} Define $f:\R\to\R$ by $f(x)=0$. Then $f\in C(\R,\R)$. Let $U$ be an open neighborhood of $f$ in $C(\R,\R)$. Assume, for a contradiction, that $\barU$ is compact. Choose $\epsilon>0$ and a compact subset $K\subseteq\R$ such that $$U_0:=\{g\in C(\R,\R)\,|\,|g|<\epsilon\hbox{ on }K\}\subseteq U.$$ (It is an unassigned exercise to prove existence of such an $\epsilon$ and $K$.) For all integers $j\ge1$, let $g_j:\R\to\R$ be defined by $g_j(x)=[\epsilon/2][\sin(jx)]$; then $g_j\in U_0$, so $g_j\in U$. We leave it as an unassigned exercise to show that $g_j$ has no convergent subsequence in $C(\R,\R)$. After all, to what could these functions possibly converge? (Recall that, in any compact Hausdorff topological space, any sequence has a convergent subsequence.) {\bf QED} \vskip.1in\noindent {\it Fact.} Let $X$ and $Y$ be locally compact topological spaces. For all $f\in C(X,C(Y,Z))$, define $\alpha_f\in C(X\times Y,Z)$ by $\alpha_f(x,y)=(f(x))(y)$. Then $f\mapsto\alpha_f:C(X,C(Y,Z))\to C(X\times Y,Z)$ is a homeomorphism. \vskip.1in If $X$ is a locally compact topological space and if $(Y,d)$ is a metric space, then the {\bf $d$-uniform on compacta topology} on $\Hom(X,Y)$ is the topology generated by the basis $\{f\in\Hom(X,Y)\,|\,\forall k\in K, d(f(k),f_0(k))<\epsilon\}$, where $f_0$ ranges over $\Hom(X,Y)$, where $K$ ranges over compact subsets of $X$ and where $\epsilon$ ranges over the positive real numbers. It is a fact that, for the topology on $Y$ generated by the basis $$\{\hbox{open balls with respect to the metric }d\},$$ the compact-open topology on $\Hom(X,Y)$ agrees with the $d$-uniform on compacta topology on $\Hom(X,Y)$. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space and let $I:=[0,1]$. Let $x,x'\in X$. A {\bf path in $X$ from $x$ to $x'$} is an element $\gamma\in C(I,X)$ such that both $\gamma(0)=x$ and $\gamma(1)=x'$. A {\bf loop in $X$ at $x$} is a path from $x$ to $x$. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space and let $I:=[0,1]$. Let $x_0,x_1\in X$ and let $$P_{x_0}^{x_1}:=\{\gamma\in C(I,X)\,|\,\gamma(0)=x_0,\gamma(1)=x_1\}$$ be the topological space of paths in $X$ from $x_0$ to $x_1$. (Give $P_{x_0}^{x_1}$ the inherited topology from $C(I,X)$.) Let $\gamma_0,\gamma_1\in P_{x_0}^{x_1}$. An {\bf endpoint fixed homotopy from $\gamma_0$ to $\gamma_1$} is a path in~$P_{x_0}^{x_1}$ from $\gamma_0$ to $\gamma_1$. We say that $\gamma_0$ and $\gamma_1$ are {\bf endpoint fixed homotopic} if there is an endpoint fixed homotopy from $\gamma_0$ to $\gamma_1$. \vskip.1in Following the preceding Fact, we may equivalently define an endpoint fixed homotopy from $\gamma_0$ to $\gamma_1$ to be a continuous $H:I\times I\to X$ such that \itemitem{(1)} for all $s\in I$, we have $H(s,0)=x_0$; \itemitem{(2)} for all $s\in I$, we have $H(s,1)=x_1$; \itemitem{(3)} for all $t\in I$, we have $H(0,t)=\gamma_0(t)$; and \itemitem{(4)} for all $t\in I$, we have $H(1,t)=\gamma_1(t)$. We have some point set topology exercises: \vskip.1in\noindent {\bf EXERCISE 3A:} Let $X$ be a topological space, let $Y$ be a set and let $f:X\to Y$ be a function. Let $\scro:=\{U\subseteq Y\,|\,f^{-1}(U)\hbox{ is open in }X\}$. Show that $\scro$ is the finest topology on $Y$ making $f$ continuous. \vskip.1in\noindent {\bf EXERCISE 3B:} Let $X$ be a topological space, let $S\subseteq X$ and let $x_0\in X$. Show that $x_0\in\partial S$ iff the following condition holds: For all open neighborhoods $U$ of $x_0$ in $X$, we have $U\cap S\ne\emptyset\ne U\backslash S$. \vskip.1in\noindent {\bf EXERCISE 3C:} Let $X$ be a topological space and let $S\subseteq X$. Show that $X\backslash(S^\circ)=\overline{X\backslash S}$. \vskip.1in\noindent {\bf EXERCISE 3D:} Let $X$ be a topological space and let $U$ be an open subset of $X$. Show that $(\partial U)^\circ=\emptyset$. That is, show that $\Int_X(\Bd_X(U))=\emptyset$. \vskip.1in\noindent {\it Definition.} Let $(X,x)$ be a pointed topological space. Define an equivalence relation $\sim$ on $P_x^x(X)$ by: For all $\gamma,\gamma'\in P_x^x(X)$, $\gamma\sim\gamma'$ means that $\gamma$ and $\gamma'$ are endpoint fixed homotopic, {\it i.e.}, that $P_\gamma^{\gamma'}(P_x^x(X))\ne\emptyset$. We define $\pi_1(X,x):=(P_x^x(X))/\sim$. \vskip.1in Note that this defines a functor $\pi_1:\{\hbox{pointed topological spaces}\}\to\{\hbox{sets}\}$. Eventually we'll redefine $\pi_1$ as a functor $\pi_1:\{\hbox{pointed topological spaces}\}\to\{\hbox{groups}\}$, and the current $\pi_1$ will be the compostion with that redefined $\pi_1$, followed by the forgetful functor $\{\hbox{sets}\}\to\{\hbox{groups}\}$. For the moment, however, this definition will suffice for our purposes. Recall that, in any topological space $Z$, if we define an equivalence relation $\cong$ on $Z$ by $z\cong z'$ iff $P_z^{z'}(Z)\ne\emptyset$, then the equivalence classes of $\cong$ are called {\bf path components of $Z$}. With that terminology, $\pi_1(X,x)$ is simply the set of path components in $P_x^x(X)$. \vskip.1in\noindent {\bf EXERCISE 3E:} Given an arrow $f:(X,x)\to(Y,y)$ in the category $$\{\hbox{pointed topological spaces}\},$$ define an arrow $\pi_1(f):\pi_1(X,x)\to\pi_1(Y,y)$ in $\{\hbox{sets}\}$. \vskip.1in As is typical with covariant functors, we use $f_*$ as an abbreviation for $\pi_1(f)$. (For {\it contra}variant functors, $f^*$ is typical as an abbreviation for the result of applying the functor to an arrow $f$.) Recall that we aim to prove that $\R^2$ and $\R^3$ are not homeomorphic. We already observed that it suffices to show that $X:=\R^2\backslash\{(0,0)\}$ and $Y:=\R^3\backslash\{(0,0,0)\}$ are not homeomorphic. It now suffices to show, for all $x\in X$ and all $y\in Y$, that $|\pi_1(X,x)|\ge2$, while $|\pi_1(Y,y)|=1$. \vskip.1in\noindent {\it Definition.} We say that a topological space $X$ is path connected if $X$ has only one path component, {\it i.e.}, if the following condition holds: for all $x,x'\in X$, we have $P_x^{x'}(X)\ne\emptyset$. We will argue, in Exercise 3G, that if $X$ is path connected, then, for any $x,x'\in X$, we have that $\pi_1(X,x)$ and $\pi_1(X,x')$ are bijective, {\it i.e.}, isomorphic in the category $\{\hbox{sets}\}$. Eventually this will be refined to saying that $\pi_1(X,x)$ and $\pi_1(X,x')$ are isomorphic in the category $\{\hbox{groups}\}$. \vskip.1in\noindent {\it Definition.} Let $x,x',x''\in X\in\scrts$. For all $\gamma\in P_x^{x'}(X)$, for all $\gamma'\in P_{x'}^{x''}(X)$, we define $\gamma\|\gamma'\in P_x^{x''}(X)$ by $$(\gamma\|\gamma')(t)=\cases{ \gamma(2t), &if $t\in[0,1/2]$;\cr \gamma'(2t-1), &if $t\in[1/2,1]$.\cr}$$ \vskip.1in The path $\gamma\|\gamma'$ is called the {\bf concatenation} of $\gamma$ with $\gamma'$. \vskip.1in\noindent {\it Definition.} Let $x,x'\in X\in\scrts$. For all $\gamma\in P_x^{x'}(X)$, we let $[\gamma]$ denote the endpoint fixed homotopy class of $\gamma$, {\it i.e.}, we define $$[\gamma]:=\{\delta\in P_x^{x'}(X)\,|\, P_\gamma^\delta(P_x^{x'}(X))\ne\emptyset\}.$$ \vskip.1in\noindent {\bf EXERCISE 3F:} Let $a,b,c,d\in X\in\scrts$. Let $\gamma\in P_a^b(X)$, $\delta\in P_b^c(X)$ and $\epsilon\in P_c^d(X)$. Show that $[\gamma\|(\delta\|\epsilon)]=[(\gamma\|\delta)\|\epsilon]$. \vskip.1in Exercise 3F asserts that concatenation is ``associative up to homotopy''. Note that it is not associative, {\it i.e.}, note that it frequently happens that $\gamma\|(\delta\|\epsilon)\ne(\gamma\|\delta)\|\epsilon$. \vskip.1in\noindent {\it Definition.} Let $a,b\in X\in\scrts$. For all $\gamma\in P_a^b(X)$, we define $\overleftarrow\gamma\in P_b^a(X)$ by $(\overleftarrow\gamma)(t)=\gamma(1-t)$. \vskip.1in\noindent {\bf EXERCISE 3G:} Let $a,b\in X\in\scrts$ and let $\gamma\in P_a^b(X)$. Show that $$[\delta]\mapsto[\overleftarrow\gamma\|\delta\|\gamma]: \pi_1(X,a)\to\pi_1(X,b)$$ is a {\it well-defined} bijection. \vskip.1in Part of the the preceding exercise is to show that, if $[\delta]=[\delta']$, then $[\overleftarrow\gamma\|\delta\|\gamma]= [\overleftarrow\gamma\|\delta'\|\gamma]$. This is what is mean by proving ``well-definedness''. One summarizes Exercise 3G by saying that ``$\pi_1$ is essentially independent of the basepoint''. This will even be true after redefining $\pi_1$ in a group theoretic way. We can define a category of ``pathed topological spaces'', each object of which consists of a topological space $X$, together with a path $[0,1]\to X$. Let $$\scra,\scrb:\{\hbox{pathed topological spaces}\}\to\{\hbox{sets}\}$$ be defined by $$\scra(X,\gamma)=\pi_1(X,\gamma(0)),\qquad\qquad \scrb(X,\gamma)=\pi_1(X,\gamma(1)).$$ Then the argument of Exercise 3G shows that $\scra$ and $\scrb$ are equivalent functors, and this is the key point of that exercise. So, for $x_0,x_1\in X$, the bijection $\pi_1(X,x_0)\to\pi_1(X,x_1)$ is ``natural up to choosing a path connecting the basepoints $x_0$ and $x_1$''. Note that, since $X:=\R^2\backslash\{(0,0)\}$ and $Y:=\R^3\backslash\{(0,0,0)\}$ are both path-connected, it now suffices to show that there exist $x\in X$ and $y\in Y$ such that $|\pi_1(X,x)|\ge2$ and such that $|\pi_1(Y,y)|=1$. (Before, we needed to show this for all $x\in X$, $y\in Y$, so the problem is now formally easier.) \vskip.1in\noindent {\it Definition.} Let $I:=[0,1]$. Let $X,Y\in\scrts$ and let $f,g:X\to Y$ be continuous. A {\bf homotopy} from $f$ to $g$ is a continuous map $H:I\times X\to Y$ such that $H(0,\cdot)=f$ and $H(1,\cdot)=g$. We say that $f$ is {\bf homotopic} to $g$, if there exists a homotopy from $f$ to $g$. \vskip.1in\noindent {\bf EXERCISE 3H:} Let $X,Y\in\scrts$, let $f,g:X\to Y$ be continuous and let $x\in X$. Assume that $f$ is homotopic to $g$. As usual, let $f_*:=\pi_1(f):\pi_1(X,x)\to\pi_1(Y,f(x))$ and let $g_*:=\pi_1(g):\pi_1(X,x)\to\pi_1(Y,g(x))$. Prove that there is a bijection $$b:\pi_1(Y,f(x))\to\pi_1(Y,g(x))$$ such that $b\circ f_*=g_*$. (Hint: If $H$ is a homotopy from $f$ to $g$, then $H(\cdot,x)$ is a path from $f(x)$ to $g(x)$. The needed bijection then comes from Exercise 3G.) \vskip.1in One summarizes Exercise 3H by the buzzphrase: ``Homotopic maps induce the same map on $\pi_1$.'' However, to be technically precise one needs to remember that the two maps on $\pi_1$, namely $f_*$ and $g_*$, yield different basepoints on $Y$, namely $f(x)$ and $g(x)$. A precise interpretation of this buzzphrase must take this into account. \vskip.1in\noindent {\it Definition.} Let $X,Y\in\scrts$. We say that $X$ is {\bf homotopy equivalent} to $Y$ (or that $X$ and $Y$ {\bf have the same homotopy type}) if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such that both of the following conditions hold: \itemitem{(1)} $g\circ f:X\to X$ is homotopic to $\id_X:X\to X$; and \itemitem{(2)} $f\circ g:Y\to Y$ is homotopic to $\id_Y:Y\to Y$. \vskip.1in\noindent {\bf EXERCISE 3I:} Let $X,Y\in\scrts$ and assume that both $X$ and $Y$ are path-connected. Assume that $X$ is homotopy equivalent to $Y$. Show, for any $x\in X$ and any $y\in Y$, that $\pi_1(X,x)$ is bijective to $\pi_1(Y,y)$, {\it i.e.}, that $\pi_1(X,x)$ is isomorphic to $\pi_1(Y,y)$ in the category $\{\hbox{sets}\}$. \vskip.1in Once again, the preceding exercise will strengthen to saying that $\pi_1(X,x)$ and $\pi_1(Y,y)$ are isomorphic in the category $\{\hbox{groups}\}$, once we've redefined the functor $\pi_1$. \vskip.1in\noindent {\bf EXERCISE 3J:} Let $n\ge2$ be an integer. Show that $S^{n-1}:=\{x\in\R^n\,|\,d(0,x)=1\}$ is homotopy equivalent to $Q:=\R^n\backslash\{0\}$. (Hint: Use the maps $p\mapsto p:S^{n-1}\to Q$ and $q\mapsto q/(d(0,q)):Q\to S^{n-1}$. \vskip.1in Given the last two exercises, our goal is now to show that there exist $a\in S^1$ and $b\in S^2$ such that $|\pi_1(S^1,a)|\ge2$ and such that $|\pi_1(S^2,b)|=1$. \vskip.1in\noindent {\it Definition.} Let $X$ be a topological space and $x\in X$. We say that $X$ is {\bf $x$-avoidable} if, for all $a,b\in X\backslash\{x\}$, for all $\gamma\in P_a^b(X)$, there exists $\gamma'\in P_a^b(X\backslash\{x\})$ such that $\gamma$ is endpoint fixed homotopic to $\gamma'$. \vskip.1in\noindent {\bf EXERCISE 4A:} Let $n\ge2$ be an integer. Let $U$ be an open, convex, nonempty subset of $\R^n$. Let $X$ be a topological space and assume that $X$ is homeomorphic to $U$. Show, for all $x\in X$, that $X$ is $x$-avoidable. (Hint: Start by showing, for all $a,b\in U$, for all $\gamma,\gamma'\in P_a^b(U)$, that $\gamma$ is endpoint fixed homotopic to $\gamma'$.) \vskip.1in\noindent {\it Lemma.} Let $X$ be a topological space and let $x\in X$. Assume that $\{x\}$ is a closed subset of $X$. Let $U$ be an open neighborhood of $x$ in $X$. Assume that $U$ is $x$-avoidable. Then $X$ is $x$-avoidable. \vskip.1in\noindent {\it Proof:} Let $a,b\in X\backslash\{x\}$ and let $\gamma\in P_a^b(X)$. We wish to show, for some $\gamma\in P_a^b(X\backslash\{x\})$, that $\gamma$ is endpoint fixed homotopic to $\gamma'$. Let $V:=X\backslash\{x\}$. Then $\{U,V\}$ is an open cover of $X$. Let $\delta>0$ be a Lebesgue number for the open cover $\{\gamma^{-1}(U),\gamma^{-1}(V)\}$ of $[0,1]$. Choose $s_0,\ldots,s_n\in[0,1]$ such that $0=s_0