Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
$P$ 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
1  
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
2  
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

1 Answer 1

up vote 2 down vote accepted

One thing you can say is that Schanuel's lemma holds. Maybe that is what you want.

share|improve this answer
    
This can be upgraded (in two directions) to the long Schanuel's lemma and the result that projective resolutions of a module are all homotopy-equivalent. The latter is more or less the statement you wanted to hold, fixed. –  Mariano Suárez-Alvarez Dec 16 '12 at 11:12
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.