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!

  • $\begingroup$ Just a note to avoid any confusion: in the second sequence that is meant to be the same $C$ as in the first sequence, not some other $C'$. $\endgroup$ Commented Apr 6, 2015 at 7:21

1 Answer 1


Schanuel’s Lemma. In the book "An Introduction to Homological Algebra" by Rotman, You can find this:

enter image description here

Remark. There exists also a dual for Schanuel’s Lemma. For this, see Exercise 3.14 of the book

  • $\begingroup$ That would be exactly what I was looking for. I figured I was close (here $\beta$ would be what I called "h" and $\alpha$ would be my $k$), I just didn't see where to go from there. Thanks! $\endgroup$ Commented Apr 6, 2015 at 7:39
  • $\begingroup$ you are welcome $\endgroup$
    – user 1
    Commented Apr 6, 2015 at 7:40

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