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:

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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.

  • $\begingroup$ 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 $\endgroup$ Commented Mar 10, 2021 at 18:30
  • $\begingroup$ 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 $X$. $\endgroup$ Commented Mar 10, 2021 at 20:36
  • $\begingroup$ 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 $\oplus$ $\endgroup$ Commented Mar 10, 2021 at 20:50
  • $\begingroup$ 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. $\endgroup$ Commented Mar 10, 2021 at 21:07
  • $\begingroup$ @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 $\endgroup$ Commented Mar 11, 2021 at 10:38

1 Answer 1


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.


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