I want mainly to ask the two question while reading Theorem 2.27 the Hatcher's textbook
Theorem 2.27) The homomorphisms between $H^{\Delta}_n(X,A) \to H_n(X,A) $ are isomorphism for all n and all $\Delta$-complex pairs $(X,A)$
one is the question about the proof of isomorphism between simplicial and singular homology group, and next question is the meaning of the proof.
To begin with, let $X^k$ be a $k$-skeleton of $X$. And then, I do not precisely explain why the following
The simplicial chain group $\Delta_n (X^k, X^{k-1})$ is zero for $n\neq k$, and is free abelian with basis the $k$-simplices of $X$ when $n = k$. ........ (*)
holds. To begin with, my idea why (*) holds :
First pick $n=k$. Then, $\Delta_k (X^{k-1})=0$ when considering the dimension. Next, k-skeleton is constructed by the costructed by attaching $n$-cells $\left \{ e_{\alpha}^n \right \} $ which is regarded as basis so that $\Delta_k (X^k)$ is a free abelian group. for convenience, I denoted by $G_{free}$ Hence, $\Delta_k (X^k, X^{k-1})$=$\Delta_k (X^k, X^{k-1})= {\Delta_k (X^k)}/ {\Delta_k (X^{k-1})} $ = $G_{free}/0 \simeq G_{free} $ . Next, since,
$$ \Delta_n(X^k)\overset{?}{=} \left\{\begin{matrix} G_{free} ~~~if~~~ n=k \\ 0 ~~~if~~~ n \neq k \end{matrix}\right.$ ...... (♣) $$
Thus,
$$ \Delta_{k-1} (X^k, X^{k-1})= {\Delta_{k-1} (X^k)}/ {\Delta_{k-1} (X^{k-1})} = 0/{G}_{free} \simeq 0 $$
and,
$$ \Delta_{i} (X^k, X^{k-1})= {\Delta_{i} (X^k)}/ {\Delta_{i-1} (X^{k-1})} = 0/0 \simeq 0 $$ if $ 0 \le i<k-1 $
Here is the first question : Is it $(♣)$ right? If $(♣)$ is wrong, I want to ask why (*) holds. Although $(♣)$ is right way, I am not confident why $(♣)$ holds.
Next question comes from the following fact that one can deduce from Theorem2.27). According to the textbook,
"We can deduce from this theorem that $H_n(X)$ is finitely generated whenever $X$ is a $\Delta$-complex with finitely many $n$ simplices, since in this case the simplicial chain group $\Delta_n(X)$ is finitely generated, hence also its subgroup of cycles and therefore also the latter group’s quotient $H^{\Delta}_n(X)$. If we write $H_n(X)$ as the direct sum of cyclic groups, then the number of $\mathbb{Z}$ summands is known traditionally as the $n$-th Betti number of $X$ , and integers specifying the orders of the finite cyclic summands are called torsion coefficients."
This description sounds that $H_n(X)$ is actually a finitely generated abelian group due to the Theorem2.27), . I think that $H^{\Delta}_n(X) $ is a finitely generated abelian group because the free abelian group $\Delta_n(X)$ which is homomorphic to finite copies of $\mathbb{Z}$ . In case of singuar homology $C_n(X)$, however, I think that $C_n(X)$ is just free abelian group which is homomorphic to $\bigoplus \mathbb{Z} $ ( the number of copies of $\mathbb{Z}$ is countable(finite or denumerable)).
Hence, my thought (or second question) is here : we could not bet that n-th singular homology group $H_n(X)$ is finitely generated without Theorem2.27 when we consider a case when the copies of $\mathbb{Z}$ is countable
Is it right thought?
Edit : When I think about the first question, my thought is wrong when considering $X^2=T$(Torus).
Like the above figure, a torus $T$ is obtained from $X^2 = X^1 \coprod_{\alpha} e_{\alpha}/\sim $ (and in fact $X^1$ is homeomorphic to $ S^1 \vee S^1 $ In that case,
$$ \Delta_{1} (X^2, X^{1})= {\Delta_{1} (X^2)}/ {\Delta_{1} (X^{1})} = <a,b,c>/<a,b,c> \simeq 0 $$
From the above, I just guess
$$ \Delta_{i} (X^k) \simeq \Delta_{i} (X^{k-1}) $$..... (◆) if $0\le i<k-1$.
Thus that $ \Delta_{n} (X^k, X^{k-1}) = 0 $ Here is my thought the relation between $\Delta_n(X^k)$ and $\Delta_n(X^{k-1})$ and I cannot generalize the case.
Answer about the question : about the question about comparing $\Delta_i(X^k)$ to $\Delta_i(X^{k-1})$, I cannot understand precise intention about the question. Clearly $X^k$ is constructed from $X^{k-1}$ by attaching $e_{\alpha}^n$. If $i=k$, $\Delta_{k}(X^{k})$ is free abelian group generated by $e_{\alpha}^n$ while $\Delta_k(X^{k-1})=0$ because of the dimension. If $i=k-1$, $\Delta_{k-1}(X^k)$ is obtained from by mapping $\partial$ : $\Delta_{k}(X^k) \to \Delta_{k-1}(X^k)$, i.e $\Delta_{k-1}(X^k)$ is means the boundary of $X^k$ while $\Delta_{k-1}(X^{k-1})$ is free abelian group generated by $e_{\alpha}^n$. If my guess ,(◆), on the previous Edit is right, these two are generated by same basis but I cannot be sure now. That's all I can do I compare them.