# Computational Homology used to verify that two spaces are homeomorphic

The following two facts seem really intriguing and I am trying to figure out how to use computations of homology to deduce them:

(a) The boundary of an ($n+1$)-simplex is homeomorphic to $S^n$.

(b) An $n$-dimensional convex body is a compact convex set in $\mathbb{R}^n$. Show that any two $n$-dimensional convex bodies are homeomorphic.

Any input to help me think about these exercises would be greatly appreciated.

-
I don't see how you could use homology to do either. Homology cannot prove the existence of homeomorphisms in general; it can only prove the non-existence of homeomorphisms. Both statements are true, however, and are easy exercises in basic topology. No input from algebraic topology is required. – Zhen Lin Feb 17 '12 at 17:27
Oh, okay. I was led to think homology should factor in, since these exercises were interlaced with several homology exercises on a problem set from a graduate course. But upon reading through the set again, I noticed that the problems are quite varied. – Vulcan Feb 17 '12 at 20:20

You can find proofs of both facts in Ch.1 section 16 of Bredon's book Topology and Geometry. Here are the two relevant statements.

16.3 Proposition. Let $C\subset \mathbb R^n$ be a compact convex body with $0\in int(C)$. Then the function $f\colon\partial C\to S^{n-1}$ given by $f(x)=x/||x||$ is a homeomorphism.

This is easy to verify now that you know what the map is!

16.4 Theorem. A compact convex body $C$ in $\mathbb R^n$ with nonempty interior is homeomorphic to the closed $n$ ball, and $\partial C\cong S^{n-1}$.

To prove this, assume by translating if necessary that $0\in int(C)$. Define $k\colon D^n\to C$ by $k(x)=||x||f^{-1}(x/||x||)$ for $x\neq 0$ and $k(0)=0$, where $f$ is as above. Now check this is a homeomorphism.

-
Not sure why the tex is rendering so poorly. – Grumpy Parsnip Feb 17 '12 at 17:56

Homological computations won't help you in proving that two spaces are homeomorphic. Having isomorphic homology groups is (usually) a much weaker condition than being homeomorphic.

In the examples you are interested in, it is possible to write down explicit homeomorphisms. My advise would be to try and find these yourself.

-