Spanier algebraic topology Proof of Lemma 5.5.9 The statement is

Let $C$ be a free chain complex such that $H(C)$ is of finite type. Then there is a free chain complex $C'$ of finite type chain equivalent to $C$. Here, a graded module $\{C_q\}$ is of finite type means $C_q$ is finitely generated for every $q$.



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*In the middle of the proof, it says '... choose for each $q$ a homomorphism $\varphi_q:F'_q\to C_{q+1}$ such that $\partial_{q+1}\varphi_q(c') = c'$. I think it uses a splitting lemma for this s.e.s.
$$0\to\ker\partial_{q+1}\to C_{q+1}\xrightarrow{\partial_{q+1}}F'_q\to 0$$
But I think $F'_q\subset\operatorname{Im}\partial_{q+1}$ but to define the above sequence, it $\operatorname{Im}(\partial_{q+1})\subset F_q'$. I'm not sure if this is true since $F'_q$ is a submodule of $Z_q(C)$.

*After that it defines a chain map $\tau$. Is the fact that $\tau$ induces an isomorphism in homology immediate? I can't see.

For injectivity, need to show for a cycle $[(c,c')]$, $[\tau(c,c')]=0$ implies $[(c,c')]=0$. Since $[(c,c')]$ is a cycle, $\partial_q'(c,c') =(c',0) = 0$ so $c' =0$. Hence, $[(c,c')] = [(c,0)]$. Now $[\tau(c,0)] = [c] = 0\in H_q(C)$ so that $c$ is homologous to an element of $F_q$, say $[c] = [b]$ for some $b\in F_q$. Since $[b] =0$, $b\in F'_q$. Hence, $[(c,c')] = [(b,0)] = 0\in H_q(C')$.
 A: *

*No, it uses the fact that $C_{q+1} \xrightarrow{\partial_{q+1}} C_q$ has image containing $F_q'$ (true by the assumption that every element of $F_q'$ is a cycle which is zero in homology) and the fact that $F_q'$ is free (it's a subgroup of a finitely generated free abelian group).

Choose a basis of $F_q'$ and choose the element $\varphi_q(c')$ whose boundary is $c'$ for each basis element, and extend linearly.



*Yes, did you try to compute?

Surjective: If $[c] \in H_q(C)$ then $c$ is homologous to an element of $F_q$ --- so $c = a + dx$ for $a \in F_q$ --- by the assumption that $F_q \to H_q$ is surjective. Therefore $[\tau(c,0)] = [c]$.
Injective: if $[(c,c')] = 0$ in $H_q(C')$ then $(c,c') = \partial_{q+1}' (x, x')$ for appropriate $x,x'$; looking at the formula this means $c = x'$ and $c' = 0$.
So if $[(c,c')] = 0$ then $(c,c') = (x',0)$. Now $\tau(x',0) = x' \in F_q' \subset C_q$, and by assumption $x' \in F_q'$ means that $x'$ is zero in $H_q(C)$, so $[\tau(c,c')] = 0$. Thus $\tau$ is injective in homology. QED.
