# Proof of acyclic models theorem in Vick's Introduction to Algebraic Topology

I am looking at the proof of the Acyclic Models theorem (5.3) in the book by James W. Vick: Homology Theory: an Introduction to Algebraic Topology.

My main issue is with the diagram on page 126:

He claims that the front face commutes due to the other faces commuting. First of all I am confused about what he means by front face. Intuitively, I would believe that the front face consists of the images of $$M_\alpha$$, but the context leads me to believe that it is actually the images of $$X$$ he is talking about when referring to the front face. Secondly, if the face with the $$X$$'s is indeed the one he means, I do not understand why commutativity of this face follows from commutativity of the others.

I will now give context about the diagram, but before reading this, I suggest having a look at the book yourself if you have the means to, since this will convey it more clearly.

• $$T$$ and $$T'$$ are functors from a certain category $$\mathcal{C}$$ to the category of chain complexes. $$T$$ is free w.r.t. a set of models of $$\mathcal{C}$$.
• $$X$$ and $$M_\alpha$$ are objects in $$\mathcal{C}$$ and more specifically, $$M_\alpha$$ comes from the set of models of $$\mathcal{C}$$.
• $$T_0$$ and $$T_0'$$ are the corresponding functors to the category of abelian groups, defined as $$(T(X))_0$$ and $$(T'(X))_0$$ respectively.
• $$H_0$$ is a functor which maps a chain complex to its zeroth homology group.
• All vertical maps are different forms of $$\pi$$, a surjective map. More specifically, the quotient map from the zeroth cycle group of a chain complex into the zeroth homology group. Due to the nature of the functors $$T$$ and $$T'$$, these cycle groups are the same as the ambient group.
• The diagonal maps are $$T_0(f),T'_0(f), H_0(T(f)), H_0(T'(f))$$, where $$f$$ is some map in $$\operatorname{Hom}_\mathcal{C}(M_\alpha,X)$$ whose existence is guarenteed if $$T_0(X)\neq 0$$. I do not think we can say anything on whether these maps are injective or surjective.
• $$\Phi$$ is some natural transformation between the functors $$H_0T$$ and $$H_0T'$$.
• Since $$T$$ is free w.r.t. the models, $$T_0(X)$$ will be free Abelian and generated by elements of the form $$T_0(f)(e_\alpha)$$ with $$e_\alpha$$ in some $$T_0(M_\alpha)$$ for a model $$M_\alpha$$. $$\phi$$ acts on these generators by $$\phi(T(f)(e_\alpha))=T'(f)(\phi(e_\alpha))$$ and $$\phi(e_\alpha)$$ is a choice of element in $$T'_0(M_\alpha)$$ such that $$\pi\circ \phi(e_\alpha) = \Phi\circ \pi(e_\alpha)$$.

For completion's sake I add the definition of being free w.r.t. models:

Fix a category $$\mathcal{C}$$ and let $$\mathcal{M}=\{M_\lambda\}_{\lambda\in\Lambda}$$ be a collection of objects in $$\mathcal{C}$$. We call this collection the models of $$\mathcal{C}$$. A functor $$T$$ from $$\mathcal{C}$$ to the category of abelian groups is free with respect to the models $$\mathcal{M}$$ if there exists an element $$e_\lambda\in T(M_\lambda)$$ for each $$\lambda$$ such that for every object $$X$$ in $$\mathcal{C}$$ $$\{T(f)(e_\lambda)\mid \lambda\in\Lambda, f\in\operatorname{Hom}_\mathcal{C}(M_\lambda,X)\}$$ is a basis for $$T(X)$$ as a free abelian group.

So from this context I believe it is clear, he can only be talking about the face consisting of $$X$$'s as the front face, since if this is commutative, then the face consisting of $$M_\alpha$$'s will be commutative aswell, seeing as $$M_\alpha$$ is also an object of $$X$$.

I believe I have shown that all faces are commutative except for the face with the $$X$$'s. So my question is how does commutativity of the whole diagram follow from this.

• This is just a guess, mainly because I didn’t work through the details here yet, but if it is about the $X$-face, it may have something to do with the fact that the $M_\alpha$ generate the $X$s. I think the maps $p_\alpha: M_\alpha \rightarrow X$ are then jointly epimorphic. By this I mean: If for all $\alpha$ the composites $fp_\alpha = gp\alpha$ agree, then $f=g$. In particular the $X$face commutes iff for all $\alpha$ the rest of the cube commutes Commented Mar 10, 2021 at 18:30
• Very interesting, the second part of your statement does hold, so if these maps $p_\alpha$ are jointly epimorphic then we are done. Could you perhaps elaborate a bit on the idea you have. I am unsure on how to have this freeness on the abelian groups mean anything on the object $X$. Commented Mar 10, 2021 at 20:36
• What I wanted to say with the first sentence is that I am not completely sure what you mean with $X$ don’t you mean $T(X)$?) is generated by $T(f)(e_\alpha)$ for some $e_\alpha \in M_\alpha$ (don’t you mean $T(M_\alpha)$. I suppose this means that being generated in particular means that there is an epimorphism $\big\oplus_\alpha T(M_\alpha) \rightarrow T(X)$. By the universal property of $\oplus$ we might write the cube either using this coproduct or for each $\alpha$ alone. The jointly epimorphic then translates into epimorphic when considering the cube with $\oplus$ Commented Mar 10, 2021 at 20:50
• Thanks for the reply. The first part of your comment is correct, I have edited the mistakes in my question. I am a bit unsure about the existence of the epimorphism you define, but I will think a bit more about this later. I have also added the definition to the question. Commented Mar 10, 2021 at 21:07
• @PrudiiArca I do not think the epimorphism you describe exists necessarily. We only have information about $e_\alpha$ and no guarantees that this fixes the whole $T(M_\alpha)$. Also, one $e_\alpha$ could belong both to $T(g)(e_\alpha)$ and $T(f)(e_\alpha)$. A counterexample which illustrates this would be the functor $S_0$ which maps a topological space to the free abelian group generated by $0$-simplices. This is modelled by $S_0(\Delta^n)$ (image of the standard simplex) with one $e_\alpha= \operatorname{Id}$. One can see that an epimorphism as you describe doesn't exist if $X$ is two points Commented Mar 11, 2021 at 10:38

Consider the family of homomorphisms $$\{T_0(f)\mid f \text{ occurs in some generator of }T_0(X)\}$$. Then this set is jointly epimorphic. Indeed, suppose we have $$g,h:T_0(X)\rightarrow Y$$ homomorphisms for which $$g\circ T_0(f) = h\circ T_0(f)$$ for all $$T_0(f)$$ in the family. For a generator $$T_0(p)(e_\alpha)$$ of $$T_0(X)$$, we then have $$g(T_0(p)(e_\alpha)) = h(T_0(p)(e_\alpha))$$ since $$T_0(p)$$ lies in the family, now since $$g$$ and $$h$$ correspond on all generators, they must be equal.
Now this is precisely our case, for any $$T_0(f)$$ in the family, we have the above 3D diagram, where a priori all but the front face (with the $$X$$'s) commute. From this it must follow that $$\Phi_X\circ\pi\circ T_0(f) = \pi'\circ\phi^0_X\circ T_0(f)$$ by going through the five commutative diagrams. But now since the family is jointly epimorphic, we see that $$\Phi_X\circ \pi= \pi'\circ \phi^0_X$$ i.e. that the front face commutes.