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For additional context (Hatcher pg. 118): Consider the pair $(X,A)$ where $A\subset X$ denotes an arbitrary subspace. The assertion from Hatcher is that $\tilde{H_0}(X,A)\cong H_0(X,A)$, with evidence coming from augmenting the SES of chain complexes$^1$ by the SES $0\to\mathbb{Z}\to\mathbb{Z}\to 0\to 0$. So it seems that Hatcher has chosen to define $\tilde{H_0}(X,A)=\frac{\ker(\epsilon)}{im(d_1)}$ where $\epsilon: Ch_0(X,A)\to 0$ is the augmentation map...? If anybody can help clarify this for me, it would be greatly appreciated. Thank you!

$^1$To provide additional clarity, we're constructing a lattice in which the SES are vertical, chain complexes are horizontal. (See photo below for visual, replacing $A_n,B_n,C_n$ with $Ch_n(A),Ch_n(X),$$Ch_n(X,A)$ resp.). The photo comes from Hatcher pg. 116).

enter image description here

J.G.131
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  • Your footnotes make no sense. There's a short exact sequence of chain complexes $0 \to C_n(A) \to C_n(X) \to C_n(X, A) \to 0$, for point 1. For point 2, the long exact sequence you get from this short exact sequence has that shape, but please don't call it a chain complex (even though it technically is), and I don't get why you cite it where you do. – Ben Steffan Apr 24 '24 at 14:31
  • @BenSteffan Sorry for being unclear. I'll update the wording in a moment. – J.G.131 Apr 24 '24 at 14:45
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    See my answers to https://math.stackexchange.com/q/3281300 and https://math.stackexchange.com/q/4098995. – Paul Frost Apr 24 '24 at 22:28

1 Answers1

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Hatcher is unfortunately not completely clear here. The idea is to define $\tilde{H}_0(X, A)$ in such a way that you have a long exact sequence ending in $\ldots \to \tilde{H}_0(A) \to \tilde{H}_0(X) \to \tilde{H}_0(X, A) \to 0$. Given that the unreduced long exact sequence is obtain from a short exact sequence $0 \to C_*(A) \to C_*(X) \to C_*(X, A) \to 0$, it is natural to ask, seeing as $\tilde{H}_*({{-}})$ is obtained from $H_*({{-}})$ by augmenting the $C_*({{-}})$, whether one can do this by "augmenting" the whole short exact sequence, i.e. whether one can tack on a bottom level short exact sequence $0 \to C_{-1}(A) \overset{f}{\to} C_{-1}(X) \to C_{-1}(X, A) \to 0$ where $C_{-1}(A) = C_{-1}(X) = \mathbb{Z}$ from the augmentation. But $f$ must be $\mathrm{id}_{\mathbb{Z}}$ since given a 0-simplex $x \in C_0(A)$, we must have $f(1) = f(\epsilon(x)) = \epsilon(i_0(x)) = 1$ for $i_0\colon C_0(A) \to C_0(X)$ induced by the inclusion since $(\ldots, i_1, i_0, f)$ must be (the components of) a chain map, so $C_{-1}(X, A) = \operatorname{coker} f = 0$. In other words, the only way to make things work out is if the augmentation of $C_*(X, A)$ is by 0, so the homology of $C_*(X, A)$ doesn't change.

Edit: As discussed in the comments, one should explicitly note that the definition of $\tilde{H}_*(X)$ as the homology of augmentation of $C_*(X)$ does not carry over to $\tilde{H}_*(X, A)$ and $C_*(X, A)$, although it is very tempting to assume so. Instead, the only way to make "everything work out," i.e. to obtain the reduced long exact sequence, is to define $\tilde{H}_*(X, A)$ to simply be the homology of $C_*(X, A)$ as well.

Ben Steffan
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  • This is a good answer, though it doesn't entirely hit what where I'm confused. My confusion is this: If we do choose to augment the chain complex of SES by $0\to\mathbb{Z}\to\mathbb{Z}\to 0\to 0$, then the induced LES of homology groups would become $...\to\tilde{H_n}(A)\to\tilde{H_n}(X)\to\tilde{H_n}(X,A)\to...\to\tilde{H_0}(A)\to\tilde{H_0}(X)\to H_0(X,A)\to 0$ according to my understood definition of reduced homology groups. – J.G.131 Apr 24 '24 at 15:06
  • And just to quickly clarify my understanding of the definition of the reduced homology groups associated with $(X,A)$: $\tilde{H_n}(X,A)$ is defined to be the $(n+1)$-homology group of the chain complex $...\to Ch_n(X,A)\to ...\to Ch_0(X,A)\to 0$ augmented specifically by $\mathbb{Z}$ with boundary map $\epsilon: Ch_0(X,A)\to\mathbb{Z}$ defined as per usual. – J.G.131 Apr 24 '24 at 15:09
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    @JAG131 Ah, that's the issue, because there's no such definition (not in Hatcher, anyway). You do have to treat $C_({{-}})$ and $C_({{-}}, {{-}})$ differently. As I tried to explain in my answer, the question here is to define what $\tilde{H}_0(X, A)$ should mean, and the "correct" way to do this is to ask: What properties do I want from this? ...and the "correct" answer is "this long exact sequence," so you define it in the only possible way that makes this sequence exist. :) – Ben Steffan Apr 24 '24 at 15:13
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    I see. That does it for me: All I needed to have clarified! Thanks again! (For your help on this post, and my post earlier. :) ) – J.G.131 Apr 24 '24 at 15:17
  • @JAG131 Glad to help! :) – Ben Steffan Apr 24 '24 at 15:18
  • Ben, If you want, I can mark your response as the “Accepted Answer” if you can quickly add that comment clarifying the definition to what exactly $\tilde{H}_{n}(X,A)$ is to the bottom of your post. – J.G.131 May 01 '24 at 00:40
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    @JAG131 There you go. – Ben Steffan May 01 '24 at 13:13