Friday, February 11, 2011

Nullhomotopy on Spheres

Quotient spaces are important concepts in (algebraic) topology. But they are also difficult to deal with. The best way to deal with them is to develop an intuition about quotient spaces. I came up with an exercise (inspired a little by Munkres) that I think illustrates the idea of quotient spaces well. It is both intuitive and formal. I hope any mathematics student who wants to become more comfortable with quotient spaces will find this helpful.

Let $n\geq 1$ be a positive integer. Suppose that $f:S^n\to Y$ is a continous function from the $n$-dimensional sphere to some arbitrary topological space $Y$ that can be extended (continously of course) to the $n+1$ dimensional (closed) disk $D^{n+1}$, that is, we can extend the domain of $f$ so that we have $f:D^{n+1}\to Y$ and $f$ restricted to the boundary of $D^{n+1}$ (which is $S^n$) is the original $f$ we started with. Then $f$ is nullhomotopic. The reason for this is very simple. The space $D^{n+1}$ is contractible to a point.

The converse statement is harder to prove, but not too hard, and we will illustrate how to use quotient spaces in proving that $f$ can be extended to a continous function on the whole $n+1$-dimensional disk if it is nullhomotopic.

The intuition here is that the sphere $S^n$ is hollow inside. It has no hole, but it is empty, and somehow if a map is homotopic to a constant map then such a map can be extended to all of region within the sphere. (The $n$-sphere is not contractible to a point, it is intuitively obvious, but just in case you do not trust your intution in higher dimensional spaces, notice that the homology groups of the sphere and a point are not the same, so clearly a sphere cannot be contracted to a point).

By assumption $f:S^n\to Y$ is nullhomotopic so there is a homotopy $F:S^n\times I\to Y$ such that $F_0(s)=f(s)$ for all $s\in S^n$ and $F_1(s)=y$ for some $y\in Y$, we will denote this homotopy as $F_t(s)$. Consider the space $S^n\times I$. You should think of this as a cylinder. To visualize this consider $S^1$ then $S^1\times I$ is really a cyclinder. But in general $S^n\times I$ should be thought as some sort of high dimensional cyclinder. The homotopy $F$ maps the upper face of the cylinder all to the same point $y$ i.e. $F(S\times 1)=y$.

The intuition now is that if we collapse (identify) the top face of the cylinder to one point then we will have defined a continous function on a cone. Formally, we construct the set $X=S^n\times I/S^n\times 1$. Let $\pi: S^n\times I\to X$ be the quotient map, we topologize $X$ by saying $U\subseteq X$ is open if and only if $\pi^{-1}(U)$ is open. Hence, we have constructed a topological space (intutively thought as a cone) $X$.

Let us get a sense of what $X$ is, just as a set. It will consist of points $\{(s,t)\}$ for all $0\leq t<1$ (we are slighly abusing notation here, $(s,t)$ is not really a pair, but an $n+2$ coordinate in $\mathbb{R}^{n+2}$ because $s$ is itself an $n+1$ coordinate) and it will consist of the point $\{S^n\times 1\}$. A cone is homeomorphic to a disk, by a simple projection operator. That is, define $p:X\to D^{n+1}$ by $p(\{(s,t)\}) = (1-t)s$ (remember $s$ is an $n+1$ coordinate so $(1-t)s$ is just a scalar product) for $0\leq t <1$. And set $p( \{ S^n\times 1\} ) = 0$. Check that $p$ is well-defined and then check that $p$ is also continous. Proving that $p$ is continous is easy but tedious so I will not do it. For example, if $U$ is open in $D^{n+1}$ and it does not contain $0$ (as an $n+1$ coordinate in $D^{n+1}$) then $p^{-1}(U)$ will be set of points $\{(s,t)\}$ in $X$ which have the condition that $s(1-t)\in U$. To show that such set of points is open in $X$ pull them back under $\pi$ (the quotient map) and check they are open in $S^n\times I$. Then consider the case when $0\in U$ and $U$ is open in $D^{n+1}$. It is straightforward but tedious so I will not do it. I think it is more important to have intutition that will convince you that $p$ must be continous. Cleary, $p$ is also a bijection. To prove that $p$ is a homeomorphism we will need to show $p^{-1}:D^{n+1}\to X$ is continous also. But we will use the following useful trick. Since $p:X\to D^{n+1}$ is a bijective continous function from a compact space to a Hausdorff space it forces $p$ to be continous (this is true for general topological spaces).

Now we will use the "univeral property" of a quotient map. So far we know that $F:S^n\times I \to Y$ is a homotopy. We also know that $\pi:S^n\times I\to X$ is a quotient map. So we can "factor" $\pi$ through a continous function $g:X\to Y$ such that $F=g\pi$. Remember that $F(s,0)=f(s)$, so it follows that $g(\pi(s,0))=f(s)$ that is $g(\{(s,0)\})=f(s)$. Finally consider the composition $gp^{-1}:D^{n+1}\to Y$. This is continous. We claim it has the desired property that $gp^{-1}$ on the boundary is the original $f$. To see this write a point in $D^{n+1}$ as $s(1-t)$. On the boundary of $D^{n+1}$ we have $t=0$, so under $p^{-1}$ the image is $\{(s,0)\}$. But by what just said $g(\{(s,0)\})=f(s)$. Therefore, we proved that if we set $h=gp^{-1}$ then $h:D^{n+1}\to Y$ is a continous function with the property that $h|_{\partial D^{n+1}} = f$. And with that we proved the theorem.