Simplicial homology of subdivision of a simplex is trivial?

Question

I've started learning a bit about simplicial homology, recently. On the way to proving topological invariance, the book ('Elements of Algebraic Topology', Munkres) uses subdivisions and simplicial approximation. However the following simple statement is taken for granted in a proof, and I haven't been able to rigorously prove it myself. Namely

If $K$ is any finite subdivision of $\Delta^n$, where $\Delta^n$ is the standard $n$-dimensional simplex, then the reduced homology $\tilde H_p(K)$ of $K$ vanishes identically.

This is used in the proof of the existence of the subdivision operator (Theorem 17.2, on page 96 of my edition). I think it might be relatively easy to see... but I'm at a loss, nevertheless.

At this point in the book, the reduced homology of a simplex and its boundary have been computed. So the above is known to be true for $K=\Delta^n$. More generally, it has been proven that a cone always has trivial homology (if that's of any help). No other homology theory (like singular homology) has been introduced, yet.

I'd very much appreciate your help! Thank you.

Answer 1

For the record, $K'(\sigma_\tau)$ is acyclic by the standing assumption of part 1: namely, that for any simplex $\sigma$ of K, the induced subdivision $K'(\sigma)$ of $\sigma$ is acyclic.

Answer 2

It might be worth pointing out that the question above is answered in part 4 of the same proof given in Munkres' book.

Sketch: Let $L = \Delta^n$ and suppose we are give a subdivision $L'$ of $L$. Let $g: L' \to L$ be a simplicial approximation to the identity on $L$.

By successive barycentric subdivision of $L$, we can find a simplicial approximation $f: \operatorname{sd}^N L \to L'$ to the identitiy on $L'$ and then, by successive barycentric subdivision of $L'$, we also find a simplicial approximation $k: \operatorname{sd}^M L' \to \operatorname{sd}^N L$ to the identity on $\operatorname{sd}^N L$. This gives us a sequence of maps

$$\operatorname{sd}^M L' \overset k\longrightarrow \operatorname{sd}^N L \overset f\longrightarrow L' \overset g\longrightarrow L$$

where $f\circ k: \operatorname{sd}^M L'\to L'$ and $g\circ f: \operatorname{sd}^N L\to L$ are simplicial approximations to the identity. Now using that a cone has trivial homology, we can see that the homology of the barycentric subdivision of a complex is the same as that of the complex. (indeed we can find an explicit subdivision operator in this case$

But then $f_\ast\circ h_\ast = 1_\ast$ and $g_\ast \circ f_\ast = 1_\ast$ implies that $f_\ast$ is an isomorphism, hence $$\tilde H_p(L') = \tilde H_p(\operatorname{sd}^N L) = \tilde H_p(L) = 0.$$

Answer 3

If ${dim}\left({\sigma}\right)$ is greater than $ p$, then any $p-$ cycle of ${sd^n}\left({\sigma}\right)$ can be pushed off the simplices not in ${sd}\left({\sigma}\right)$ by subtracting a suitable p-boundary ,the trick munkres uses in sec. 5 of this book.Then as ${sd}\left({\sigma}\right)$ is acyclic being a cone we're done.If ${dim}\left({\sigma}\right)$ = $p$,then for any $p-$ cycle $c$, the boundary , $\partial{c}$= $0$, so c must be equal on all simplices and hence a multiple of $\sum_{i}{\tau}_{i}$ ( where $\tau_{i}$ ranges over all $p$ -simplices of $\sigma$) ,but then since it is $0$ on $Bd(\sigma)$ it must be $0$ all over. So , ${sd^n}\left({\sigma}\right)$ is acyclic.