# Cancellation law of morphisms?

The title is probably misleading, but I wasn't quite sure how to boil my question down to one line.

My problem is this:

Assume I have two projective $R$-modules, $P$ and $Q$, and epimorphisms $\varphi: P \to M$ and $\psi : Q \to M$, where $M$ is any $R$-module. Since $P$ and $Q$ are projective, I know we have $f: P \to Q$ and $g: Q \to P$ such that $$\varphi = \varphi \circ (g \circ f) \text{ and } \psi = \psi \circ (f \circ g),$$ but what does that tell me about the relationship between $P$ and $Q$?

I mean loosely speaking, I can see that with respect to some element $m$ of $M$, the "class" of elements that maps to $m$ from $P$ are isomorphic to the "class" of elements that maps to $m$ from $Q$ ($\varphi^{-1}(m) \cong \psi^{-1}(m)$ for all $m \in M$), but isn't this just the canonical isomorphism $P/\ker{\varphi} \cong M \cong Q / \ker{\psi}$ (or am I wrong here?), we get from extending $P \to M \to 0$ to the obvious short exact sequence $0 \to \ker{\varphi} \to P \to M \to 0$.

Can I use the fact that $P$ and $Q$ are projective to say anything more specific about the relation between $P$ and $Q$ (for example that they are isomorphic)? Or perhaps about the relation between the kernels $\ker{\varphi}$ and $\ker{\psi}$?

Thanks!

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$P$ and $Q$ need not be isomorphic, the situation is different to universal properties where a unique morphism in place of $f,g$ here would result. You will get uniqueness/well-definedness again only at the homological level. – Hagen von Eitzen Dec 16 '12 at 11:03
In fact, you should have no trouble coming up with exampes where $P$ and $Q$ are not isomorphic! – Mariano Suárez-Alvarez Dec 16 '12 at 11:05
Always, always, always look at examples. To go to the extreme, see what happens if you take $M=0$ here! – Mariano Suárez-Alvarez Dec 16 '12 at 11:08

Let me see if I got this straight: What I do get is that $P \oplus \ker{\psi} \cong Q \oplus \ker{\varphi}$? It just feels like I should get something "stronger" than that (or maybe I just don't realize the full imlpications of this result yet :) ), but it feels like I ought to be able to say more... – user48168 Dec 16 '12 at 11:18
Ah, but of course... This holds for any two projective modules and any $M$ with two such epimorphisms described, so it couldn't be too strong an implication. I guess I'll sit down and look at some examples a little more, untill I start to get a better grasp of it... – user48168 Dec 16 '12 at 11:24