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Q1) In the paper ([Zomorodian, page 6 https://pdfs.semanticscholar.org/b7ae/132ef88bde28903ac8b14a29a76130f61ac2.pdf), it is stated without proof that $im\ \eta_k^{i,p}\cong H_k^{i,p}$. May I know how do we prove it?

Q2) Also, it states that $\eta_k^{i,p}: H_k^i\to H_k^{i+p}$ is an injection that maps a homology class into the one that contains it. Is the word "contains it" meant to be literal as in set containment, or is it just a word to reflect the injectivity?

Thanks.

Some background of the terminology used (feel free to ask me for clarifications):

$H_k^{i,p}=Z_k^i/(B_k^{i+p}\cap Z_k^i)$ where $Z_k^i=\ker(\partial_k^i)$ is the cycle group of the $i$th filtered complex $K^i$, and $B_k^i=im\ \partial_{k+1}^i$ is the boundary group.

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I don't think that $\eta_k^{i,p}: H^i_k \to H^{i+p}_k$ is an injection in general (but then again I have not read the paper carefully enough to see whether they impose additional conditions to make it so).

Anyway, for (1), observe that by the first isomorphism theorem, we have $$\operatorname{im} \eta_k^{i,p} \cong H^i_k/\ker \eta_k^{i,p}.$$ By definition, $H^i_k := Z^i_k/B^i_k$. What about $\ker \eta_k^{i,p}$? It consists of $k$-cycles in the $i$-th filtered complex ($Z^i_k$) that become $k$-boundaries in the $(i+p)$-th filtered complex ($B^{i+p}_k$), modulo $k$-boundaries in the $i$-th filtered complex ($B^i_k$). In other words, $$\ker \eta_k^{i,p} \cong (Z^i_k \cap B^{i+p}_k)/B^i_k.$$

Putting this together and using the third isomorphism theorem, we have $$\operatorname{im} \eta_k^{i,p} \cong \frac{Z^i_k/B^i_k}{(Z^i_k \cap B^{i+p}_k)/B^i_k} \cong \frac{Z^i_k}{Z^i_k \cap B^{i+p}_k} =: H^{i,p}_k$$ as claimed.

For (2), I don't think there's a natural and useful way to compare $H^i_k \to H^{i+p}_k$ by set-theoretic containment. But they are quotient groups of things that can be compared. To wit, let $[x] \in H^i_k$ be a homology class. Here's what the map $\eta_k^{i,p}$ does. Let $x \in Z^i_k$ be a cycle representing the homology class $[x]$, which we can then regard as a cycle not in $K^i$, but in $K^{i+p}$, since $K^i \subseteq K^{i+p}$. Then $\eta_k^{i,p}([x])$ is the homology class of the class $x$ in $K^{i+p}$. I guess it's up to you to decide whether you think this means that "$\eta_k^{i,p}([x])$ contains $[x]$".

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  • $\begingroup$ Very nice answer. Helped me a lot, thanks! $\endgroup$ – yoyostein Jul 4 '17 at 14:53

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