If $Ext(P,\_) = 0,$ then $P$ is projective

Question

Trying to understand Bredon's proof (below). Please note that $G$ is an Abelian group.

So my confusion starts with $H_1(Hom(R,G_*)) \to H_0(Hom(A,G_*)) \to H_0(Hom(F,G_*))$. I understand that he is using Zig-Zag lemma:

$H_2(Hom(R,G_*)) \to H_2(Hom(A,G_*)) \to H_2(Hom(F,G_*)) \to H_1(Hom(R,G_*)) \to H_1(Hom(A,G_*)) \to H_1(Hom(F,G_*)) \to H_0(Hom(R,G_*)) \to H_0(Hom(A,G_*)) \to H_0(Hom(F,G_*))$.

Now because $F$ is projective $Hom(F,G_*)$ is exact, so $H_i(Hom(F,G_*))=0$.

$Hom(R,G_*)$ is only left exact, so can we find $H_i(Hom(R,G_*))$? I thought maybe $H_2(Hom(R,G_*))=0=Ext^2(R,G)$, since G is abelian and similar for $H_2(Hom(A,G_*))=0=Ext^2(A,G)$ but is this correct and then what happens to $H_0(Hom(R,G_*))$?

Answer 1

I think there's some confusion indeed: $\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Ext}{\operatorname{Ext}} \Ext^2(X, G)$ is indeed trivial for all $X$, but this has nothing to do with $H_2(\Hom(X, G_*))$ ($X = R, A$): For one, $G_*$ and $G$ are completely unrelated; there is no group by the name of $G$ in that specific part of the argument. Moreover, $G_*$ is just some short exact sequence and you're not taking homology at the right index, so the group does not really connect to $\Ext^2$ in any meaningful way.

As for how to determine these groups, there's two ways to go about it:

1. Use the long exact sequence. Note that the homology sequence you wrote down actually extends by 0 to the left and right. Let us then see what terms we can eliminate using Bredon's arguments: All $H_i(\Hom(F, G_*))$ are trivial by projectivity of $F$, and $H_1(\Hom(R, G_*)) = 0$ as mentioned as well as $H_2(\Hom(R, G_*))$ by the same argument. Now all remaining terms are wedged between two zeroes each, hence all trivial.
2. Invoke some knowledge about $\mathbb{Z}$-modules. Namely, a $\mathbb{Z}$-module (which is just another name for abelian group) is free iff it is projective, and submodules of free modules are again free, so $F$ is free, $R$ is free as a submodule of $F$, and if you accept the conclusion of the proof then $A$, too, is free, from which it follows that $\Hom(X, G_*)$ is exact for $X = A, F$ and/or $R$, whence all homology of these complexes vanishes.