Proof of the acyclic model theorem - having trouble I am using as a reference Spanier's text Algebraic Topology.
The acyclic model theorem is stated as theorem 5.3.3. 
I am unable to show that the natural transformation $H_0G\to H_0G'$ is well-defined on equivalence classes. For example, if $g$ is a boundary in $G_0(X)$, how do I show that $\sum n_{ij}G'_0(f_{ij})z_j$ is also a boundary?
 A: First of all let us note that Spanier is little imprecise. He defines a functor $G$ from a category $\mathcal C$ with models $\mathfrak M$ to the category of chain complexes to be free if all $G_q$ are free. However, it seems to me that he really works with a collection of functors $G_q$ from $\mathcal C$ to the category of abelian groups which are free on models $\mathfrak M_q$ which may explicitly depend on $q$. In many cases these sets must even be pairwise disjoint. In Example 6, one should take $\mathfrak M_q(K) = \{ \bar s \mid s \in K, \dim s = q \}$. If we take $\mathfrak M(K) = \{ \bar s \mid s \in K \}$ for all $q$, then Spanier's $g_s = \sigma(s)$ are not contained in $C_q(\bar s)$ unless $\dim s = q$. Similarly, in Example 7 we must take $\mathfrak M_q = \{ \Delta^q \}$.
In the proof of Theorem 3 Spanier assigns to each $g_j \in G_0(M^0_j)$, $M^0_j \in\mathfrak M_0$, the unique $z_j \in H_0(G'(M^0_j))$ such that $\bar{\epsilon}([g_j])= \bar{\epsilon}'(z_j)$, where $\bar{\epsilon} : H_0(G(M_j)) \to \mathbb Z, \bar{\epsilon}' : H_0(G'(M_j)) \to \mathbb Z$ are induced by the augmentation maps, and defines $H_0(G(X)) \to H_0(G'(X))$ by
$$\left[\sum n_{ij}G_0(f_{ij})(g_j)\right] \mapsto \sum n_{ij}(f_{ij})_*(z_j) .$$
It is not clear why this is well-defined. To see this, we need a slightly modified approach. Choose $\zeta_j \in G'_0(M_j)$ such that $[\zeta_j] = z_j$ and define $\theta_0 : G_0(X) \to G'_0(X)$ by
$$\theta_0(\sum n_{ij}G_0(f_{ij})(g_j)) = \sum n_{ij}G'_0(f_{ij})(\zeta_j) .$$
Recall that the $G_0(f)(g_j)$ form a basis for $G_0(X)$, i.e. $\theta_0$ is the unique homomorphism such that $\theta_0(G_0(f)(g_j)) = G'_0(f_{ij})(\zeta_j)$. We have $\epsilon' \theta_0 = \epsilon$ because $\epsilon'G_0(f) = G'_0(f) \epsilon$ and $\epsilon(g_j) = \epsilon'(\zeta_j)$.
To see that $\theta_0$ maps boundaries to boundaries, we use the same method to construct $\theta_1 : G_1(X) \to G'_1(X)$ such that $\partial'_1\theta_1 = \theta_0 \partial_1$. This will show that Spanier's map is well-defined.
Since $\epsilon' \theta_0 \partial_1 = \epsilon \partial_1 = 0$, for each $g^1_k  \in G_1(M^1_k)$, $M^1_k \in \mathfrak M_1$, we have $\theta_0 \partial_1(g^1_k) \in \ker \epsilon'$. Since $G'$ is acyclic, we know that $\ker \epsilon' = \text{im} \partial'_1$. Hence we can choose $z^1_k \in G'_1(M^1_k)$ such that $\partial'_1(z^1_k) = \theta_0 \partial_1(g^1_k)$. Now define $\theta_1$ as $\theta_0$.
Edited:
After a little exchange of ideas with Patrick (see the comments) I suggest the following "pragmatic" definition of a free functor.
A category with models is a pair $(\mathcal C, \mathfrak M)$ consisting of a category $\mathcal C$ and a set $\mathfrak M$ of objects of $\mathcal C$.
Consider a functor $G : \mathcal C \to \mathbf{Ab}$ = category of abelian groups. An $\mathfrak M$-basis for $G$ consists of a subset $\mathfrak M_G \subset \mathfrak M$ and an element $(g_M ) \in \prod_{M  \in \mathfrak M_G} G(M)$ such that the set
$$\{G(f)(g_M) \mid M \in  \mathfrak M_G, f \in \hom(M,X) \}$$
is a basis for $G(X)$. If $G$ has an $\mathfrak M$-basis, then $G$ is called free on $(\mathcal C, \mathfrak M)$.
Edited:
In the proof of Theorem 5.3.3 Spanier refers to Theorem 5.2.8 which requires that $G'$ is acyclic in positive dimensions. The stronger assumption that $G'$ is acyclic, i.e. the additional assumption that $\tilde{H}_0(G'(M)) = H_0(\tilde{G'}(M)) = 0$ for all $M \in \mathfrak{M}$, enters when he constructs the elements $z_j \in H_0(G'(M^0_j))$ which are uniquely determined by  $\bar{\epsilon}([g_j])= \bar{\epsilon}'(z_j)$ (due to Lemma 1 and $\tilde{H}_0(G'(M_j)) = 0$). It is clear that any natural transformation $\Theta : H_0(G) \to H_0(G')$ which is induced by an augmentation preserving natural chain map must necessarily satisfy $\Theta([g_j]) = z_j$. It is then also clear that the only candidate for $\Theta : H_0(G(X)) \to H_0(G'(X))$ is that described above. I think it is a serious omission by Spanier not to clarify why this is well-defined, and it puzzles readers. The proof requires to use a second time $\tilde{H}_0(G'(M_j)) = 0$.
On the other hand, I think that Theorem 5.3.3 is completely covered by Theorem 5.2.8. Each augmented chain complex $C$ can be regarded as an ordinary nonnegative chain complex $C^{sh}$ by making a dimension shift of $+1$. Each augmentation preserving chain map $\phi : C \to D$ corresponds to an ordinary chain map $\phi^{sh} : C^{sh} \to D^{sh}$ which is the identity in dimension $0$. With this construction we have $\tilde{H}_q(C) = H_{q+1}(C^{sh})$. Clearly, if $G$ is free, then also $G^{sh}$ is free (in dimension $0$ we have the constant functor $X \mapsto \mathbb Z$ for which a basis is $1 \in \mathbb Z = const(M)$ for an arbitrary $M$). Also, if $G'$ is acyclic, then $(G')^{sh}$ is acyclic in positive dimensions. Now apply Theorem 5.2.8 to $id : H_0(G^{sh}) \to H_0((G')^{sh})$. Note that this is the only homomorphism which can be induced by an augmentation preserving natural chain map $G \to G'$ (consider the corresponding natural chain map
$G^{sh} \to (G')^{sh}$).
Last edit:
I just realized that the best approach would be to replace Theorem 5.2.8 by the following (with the same assumptions as in 5.2.8):
(a) Any natural transformation $\nu : G_0 \to H_0(G')$ is induced by a natural chain map $\tau : G \to G'$. This means that $\nu = \pi' \circ \tau_0$, where $\pi' : G'_0 \to H_0(G')$ is the natural quotient map.
(b) Two natural chain maps $\tau, \tau' : G \to G'$ inducing the same natural transformation $G_0 \to H_0(G')$ (or equivalently, the same natural transformation $H_0(G) \to H_0(G')$) are naturally chain homotopic.
The proof is - mutatis mutandis - the same as that of 5.2.8. The new theorem covers Spanier's theorem because


*

*For any natural transformation $\theta: H_0(G) \to H_0(G')$, the composition $\theta \circ \pi : G_0 \to H_0(G')$ is a natural transformation, where $\pi : G_0 \to H_0(G)$ is the natural quotient map.

*Each natural transformation $\nu : G_0 \to H_0(G')$ which is induced by a natural chain map $\tau : G \to G'$ induces a unique natural transformation $\theta: H_0(G) \to H_0(G)$ such that $\theta \circ \pi : G_0 \to H_0(G)$.
This variant of Theorem 5.2.8 can be used to prove Theorem 5.3.3.
