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For what values of $i$ is it true that $H_i(K^{(p)})\simeq H_i(K)$?

My guess is that this is true for $i>dim K$. Otherwise, we can use $n$-simplex for a counterexample.

Stefan Hamcke
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Hashirama
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2 Answers2

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If $X$ is a simplicial or more generally a CW-complex, then $H_k(X^{n}) \cong H_k(X^{n+1})$ for all $k < n$. There's a way to see this without the machinery of cellular homology.

Consider the long exact homology sequence of the pair $(X^{n+1}, X^{n})$, which is

$$\cdots \to H_{k+1}(X^{n+1}/X^{n}) \to H_k(X^{n}) \to H_k(X^{n+1}) \to H_k(X^{n+1}/X^{n}) \to \cdots$$

$X^{n+1}/X^n$ is a wedge sum of $(n+1)$-spheres, which has nontrivial homology only at dimension $n+1$. Thus $H_k(X^{n+1}/X^n) \cong H_{k+1}(X^{n+1}/X^n) \cong 0$, which leads to the result.

Thus, by induction, $H_k(X^n) \cong H_k(X)$ for all simplicial or more generally CW-complex $X$ whenever $k < n$.

(as pointed out in a comment below by @StefanHamcke, this argument works only if $X$ is finite dimensional. However, this is also true for infinite dimensional CW-complexes. For a proof, look at Hatcher's Algebraic Topology, page 138)

Stefan Hamcke
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Balarka Sen
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  • This is essentially @EricAuld's answer, but without explicitly spelling "cellular chain complex". – Balarka Sen Jan 23 '16 at 19:02
  • Unless I'm mistaken, you can't say $X^{n+1}/X^n$ is a wedge sum of $(n+1)$-spheres, only that is the wedge sum of some number of spheres. – Michael Albanese Jan 23 '16 at 23:16
  • @MichaelAlbanese By $(n+1)$-spheres, I mean wedge of $S^{n+1}$'s, not wedge of $(n+1)$ number of spheres. Obviously, number of spheres in the wedge is the number of $(n+1)$-cells in $X$. – Balarka Sen Jan 24 '16 at 06:54
  • I see. I missed the importance of the hyphen. – Michael Albanese Jan 24 '16 at 06:55
  • I don't think the final isomorphism $H_k(X^n)\cong H_k(X)$ follows that easily by induction. You have to look at a $k$-cycle $c$ in $X$ and argue with some compactness argument that $c$ is contained in some skeleton $X^m$, and then you can use induction. – Stefan Hamcke Jan 24 '16 at 19:27
  • @StefanHamcke Good point, my CW-complexes are all finite dimensional. – Balarka Sen Jan 24 '16 at 19:30
  • Well, then it works, of course. Maybe you could add that to your answer, otherwise the OP could think that induction does indeed immediately give the result. – Stefan Hamcke Jan 24 '16 at 19:35
  • @StefanHamcke I have edited this and a reference for a proof for infinite dimensional $X$ in. – Balarka Sen Jan 24 '16 at 19:38
  • Thanks, +1. I took the freedom to do some minor corrections to your post :-) – Stefan Hamcke Jan 24 '16 at 19:46
  • @StefanHamcke Yep, I noticed. Thanks for that, my dimensions were a mess. – Balarka Sen Jan 24 '16 at 19:47
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By looking at cellular homology, we can see that the $n$th homology group of a CW complex is unaffected by adding cells of dimension $n+2$ and higher. (It affects neither the kernel of $\partial_n$ nor the image of $\partial_{n+1}$.)

The same thing applies for simplicial homology, replacing "cell" with "simplex".

Eric Auld
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