Exact sequence with reduced homology I want to verify my understanding of reduced and relative homology groups, so I’d like to verify my proof to Problem 5.19 of Rotman’s Algebraic topology.
Problem 5.19

5.19.
If $A \subset X$, then there is an exact sequence
$$
  \dotsb \to \tilde{H}_n(A) \to \tilde{H}_n(X) \to \tilde{H}_n(X, A) \to \tilde{H}_{n - 1}(A) \to \dotsb \,,
$$
which ends
$$
  \dotsb \to \tilde{H}_0(A) \to \tilde{H}_0(X) \to H_0(X, A) \to 0 \,.
$$
(Hint: $\tilde{S}_*(X) / \tilde{S}_*(A) = S_*(X) / S_*(A)$.)
(Original scan)

Notation: $\tilde{H}_n(X)$ is the reduced homology group of $X$, and $\tilde{S}_*(X)$ is the augmented singular complex with generator $[\enspace]$ (I noticed that wikipedia regards $\tilde{S}_{-1}(X) = \mathbb{Z}$, which is equivalent and a little cleaner).
Proof
The inclusions $i \colon A \hookrightarrow X$ and $j \colon (X, \emptyset) \hookrightarrow (A, X)$ induce chain maps $i_\#$ and $j_\#$ such that
$$ 0 \rightarrow \tilde{S}_*(A) \xrightarrow{i_\#} \tilde{S}_*(X) \xrightarrow{j_\#} \tilde{S}_*(X, A) \rightarrow 0$$
is exact. Applying the Exact Triangle,
$$
\cdots \rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow \tilde H_0(X, A) \rightarrow 0
$$
is exact. So it remains to show that $\tilde H_n(X,A) = H_n(X,A)$ for every $n$. Note that $\tilde S_n(X) / \tilde S_n(A) = S_n(X) / S_n(A)$ for every $n \geq 0$
But $\tilde{S}_{-1}(X)$ and $\tilde{S}_{-1}(A)$ have the same generator $[\enspace]$, and so $\tilde{S}_{-1}(X) = \tilde{S}_{-1}(A)$ implies that
$$\tilde{S}_{-1}(X) / \tilde{S}_{-1}(A) = \overline 0 \cong 0 = S_{-1}(X) / S_{-1}(A) \tag{1}$$
Thus, $\tilde{S}_*(X) / \tilde{S}_*(A) \cong S_*(X) / S_*(A)$ and $\tilde{H}_n(X,A) = H_n(X,A)$ for every $n$. $\square$
Questions
Rotman’s hint says that the quotient complex for the augmented complex should equal the non-augmented one. But my equation $(1)$ suggests that they are only isomorphic. Am I mistaken here? I’m also not sure if isomorphic complexes have equal homologies?
 A: This is a question about the interpretation of the symbol $0$. Usually one understands $0$ to be the trivial (abelian) group. But what does this mean? Any group with one element deserves to be called trivial group. Of course we can pick a specific "model" for $0$, for example $0  = \{\emptyset\}$, but this is unnecessary (not to say counterproductive). Writing $G = 0$ simply means that the group $G$ has a single element $e$, and that is all we need to know about $G$; it is irrelevant what $e$ is. Here are some arguments.

*

*Let $f : G \to H$ be a group homomorphism. It is customary to write $f = 0$ if $f$ maps all elements of $G$ to the neutral element of $H$. In this case we write $\operatorname{im} f = 0$ which means that the image of $G$ under $f$ has one element (the neutral element of $H$). In other words, the subgroup $\operatorname{im} f$ of $H$ is trivial. We must not think that $\operatorname{im} f = 0$ means that $\operatorname{im} f$ is always the same trivial group independent of $H$.


*Consider a chain complex $\mathcal C = (C_n,\partial_n)$. If some $C_k$ is a trivial group, we write $C_k = 0$. In this case we have $\ker \partial_{k+1} = C_{k+1}$ and  $\operatorname{im} \partial_k = 0$, independent of what $C_k$ looks precisely. Note that kernels and images are all  what is needed for the purposes of homology. Also note that if $\operatorname{im} \partial_{k+1} = 0$ (which may happen also if $C_{k+1} \ne 0$), then $H_k(\mathcal C) = \ker \partial_k/\operatorname{im} \partial_{k+1} \approx \ker \partial_k$ via the natural quotient map $\pi_k :  \ker \partial_k \to \ker \partial_k/\operatorname{im} \partial_{k+1}$ which is a group isomorphism in this case.


*Consider an exact sequence
$$ \ldots \stackrel{f_{n+2}}{\to} G_{n+1} \stackrel{f_{n+1}}{\to} G_n \stackrel{f_{n}}{\to} G_{n-1} \stackrel{f_{n-1}}{\to} \ldots $$
If $G_{n}$ is a trivial group, we write $G_{n} = 0$. In this case exactness means that $f_{n+2}$ is surjective and $f_{n-1}$ is injective, independent of what $G_n$ looks precisely.
In your equation $(1)$ both $\tilde{S}_{-1}(X) / \tilde{S}_{-1}(A)$ and $S_{-1}(X) / S_{-1}(A)$ are trivial groups which are technically distinct. But as we have seen this does not make a difference for the purposes of homology.
