Problem: Prove that for two short exact sequences
$$ 0\rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0 $$ $$ 0\rightarrow A' \xrightarrow{f'} B' \xrightarrow{g'} C \rightarrow 0, $$
where $B$ and $B'$ are projective $R$-modules, one has
$$ B\oplus A' \cong B'\oplus A. $$
Work: So far what I've thought up is the fact that $B'$ is projective and $g$ is surjective implies the existence of a lifting $h:B\rightarrow B'$, i.e.
$$g'\circ h=g. $$
Similarly (swapping $B'$ with $B$ and $g$ with $g'$), we get a lifting $h':B'\rightarrow B$, i.e.
$$g\circ h'=g'. $$
In addition to this, for any $a\in A$ we have that
$$ g'(h(f(a)))=g(f(a))=0, $$
so $h(f(a))\in\text{ker}(g')=\text{im}(f')$; thus, we find some $a'\in A'$ such that $f'(a')=h(f(a))$. Using this, we can define a map $k:A\rightarrow A'$ by sending $a\mapsto a'$. (Note: the fact that $k$ is well-defined follows from the injectivity of $f$.) Similarly, we have a map $k':A'\rightarrow A$ characterized by
$$ g'(a')=a \quad\Leftrightarrow\quad f(a)=h'(f'(a')). $$
This leads to maps $h'\oplus k: B'\oplus A\rightarrow B\oplus A'$ and $h\oplus k': B\oplus A'\rightarrow B'\oplus A$.
Question: It is not at all clear to me that these maps should be inverses of each other. Are they? Any hints on how to show it? Or is there another way to approach this problem?
Any help would be greatly appreciated! Thanks!
Schanuel’s Lemma. In the book "An Introduction to Homological Algebra" by Rotman, You can find this:
Remark. There exists also a dual for Schanuel’s Lemma. For this, see Exercise 3.14 of the book