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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.

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  • $\begingroup$ In the version of the book I'm looking at, your statement is not assumed, and is indeed a consequence of Theorem 17.2. The proof is long, could you say a bit more about where and how this statement is used? $\endgroup$ – Justin Young Sep 21 '12 at 7:01
  • $\begingroup$ @JustinYoung: Yes, I will gladly do so! I have a paperback edition. The statement is in part 1 of the proof. The point I'm confused about is below figure 17.1 in my edition, where it is written "[...] We define $$\Theta(\tau) = K(\sigma_\tau),$$ $$\Phi(\tau) = K'(\sigma_\tau);$$ both complexes are acyclic. [...]" Here $\sigma_\tau$ is a simplex of the complex $K$, $K'$ is a subdivision of $K$ and $K'(\sigma_\tau)$ denotes the complex which is the union of all simplices of $K'$ which lie in $\sigma_\tau$. I don't see how to go about proving that $K'(\sigma_\tau)$ is acyclic. $\endgroup$ – Sam Sep 21 '12 at 19:32
  • $\begingroup$ Ah, well, $\sigma_\tau$ is a simplex in $K$, so $K(\sigma_\tau)$ is just that simplex and its faces, therefore acyclic. As for $K'(\sigma_\tau)$, that 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. $\endgroup$ – Justin Young Sep 21 '12 at 20:39
  • $\begingroup$ @JustinYoung: Ah, damn. :) Looks like I had already forgotten this standing assumption when I came to this part of the proof. Thank you very much for pointing it out. If you'd like to copy your comment as an answer, I'll happily accept it! $\endgroup$ – Sam Sep 22 '12 at 16:43
  • $\begingroup$ @SamL. Is this used on the way to prove Excision? $\endgroup$ – user38268 Sep 23 '12 at 8:59
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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.

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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 $$\tilde H_p(L') = \tilde H_p(\operatorname{sd}^N L) = \tilde H_p(L) = 0.$$

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