Horse shoe lemma exercise from Weibel's Homological Algebra

Question

This is regarding the exercise 2.2.4 in Weibel's Introduction to homological Algebra which is about Horse shoe lemma for projective resolutions. That is if $0 \to A' \to A \to A'' \to 0$ is an exact sequence of R modules and if ${P'}_{.}$ and ${P''}_{.}$ and projective resolutions of $A'$ and $A''$ respectively then there is a projective resolution ${P}_{.}$ of A such that $0 \to {P'}_{.} \to {P}_{.} \to {P''}_{.} \to 0$ is a split exact sequence of complexes. Proof I understood and the differentials for ${P}_{.}$ are constructed step by step. But in the exercise he has given an explicit form for that. I am not able to prove that. Please help me.

Answer 1

Weibel's construction is perhaps a bit obscure (although effective if you mean to prove that such resolution exists in a quick way). The idea of the horseshoe lemma, among other things, is to obtain a split resolution of a given short exact sequence of modules. Thus, you want to define an appropriate differential $d : P'\oplus P'' \longrightarrow P'\oplus P''$ that make the desired diagram commute, where the resolution of your exact sequence is the (canonically) split exact sequence of complexes

$$ P' \longrightarrow P'\oplus P''\longrightarrow P''$$

You don't need to check the middle complex is a resolution once you obtained one, because it is automatically exact by the homology LES of the last short exact sequence.

If we write $\varepsilon',\varepsilon''$ for the augmentations for $A'$ and $A''$, then writing $\varepsilon_0'+\varepsilon_0''$ for the augmentation of $A$ where $$\varepsilon_0':P_0' \longrightarrow A$$ $$\varepsilon_0'':P_0'' \longrightarrow A$$

following the arrows gives

$$\varepsilon_0' =f\varepsilon'$$ $$\varepsilon ''\pi = g\varepsilon_0''$$

and because $P_0''$ is projective you can solve for $\varepsilon_0''$. Now $d$ is a matrix of morphisms $\begin{pmatrix} f_1 & f_2 \\ g_1&g_2 \end{pmatrix}$. In particular, if we want commutativity at $P'$ we need that (follow the arrows, again)

$$f_1x'= d'x'$$ $$g_1x'=0$$

and commutativity at $P''$ gives that $g_2x''=d''x''$. We have shown that $$d=\begin{pmatrix} d' & f_2 \\ 0&d'' \end{pmatrix}$$ which is what Weibel wanted.

Thus it suffices to determine $f_2:P''\longrightarrow P'$ inductively so that $d^2=0$, or, what is the same, so that $$d'f_2 +f_2 d''=0$$ and I leave that to you. Note we already determined the map at degree $0$, where it looks a bit different. At any rate, the proof is an inductive lifting argument.