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I want to show that if projective covers exist then they are unique up to isomorphism.

More precisely let $f: P \rightarrow M$ and $g: Q \rightarrow M$ be projective covers of an $R$-module $M$.

Using the fact that $g$ is surjective and that $P$ is projective we can find an $R$-map $h: P \rightarrow Q$ such that $g \circ h=f$.

Note then that $\operatorname{Im}(h)+\operatorname{ker}(g)=Q$. Since $ker(g)$ is superfluous this implies that $Q=Im(h)$ so that $h$ is surjective.

But how do we conclude that $h$ is injective?

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I'm not sure what category you're working in, but I'm pretty certain it's not true if you're working in the category of $R$-modules. For example, take $Q = P \oplus P$. – Zhen Lin Apr 19 '12 at 7:20
@zhen, if P is a cover then that Q won't be. – Mariano Suárez-Alvarez Apr 19 '12 at 8:34
Hm? Why would $g$ be a monomorphism? (if it were, since it is also surjective, $M$ would be projective...) – Mariano Suárez-Alvarez Apr 19 '12 at 18:25
up vote 3 down vote accepted

Repeat your argument. Using the fact that $f$ is surjective and $Q$ is projective there exists a map $h'$ such that $f \circ h' = g$. As above, you conclude that $h'$ is surjective, and you have $g \circ h \circ h' =g$.

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