Clarification of terminology: We say an injection $A \xrightarrow{i} B$ splits iff the induced short exact sequence $0 \to A \xrightarrow{i} B \to B/i(A) \to 0$ splits. Similarly, we say a surjection $B \xrightarrow{p} C$ splits iff $0 \to \ker p \to B \xrightarrow{p} C \to 0$ splits.
Here are the definitions of injective and projective modules I'm using:
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$Q$ is injective if, and only if, $\operatorname{Hom}(\ast, Q)$ takes injections to surjections ($0 \to A \to B$ exact implies $\operatorname{Hom}(B,Q) \to \operatorname{Hom}(A,Q) \to 0$ exact).
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$P$ is projective if and only if $\operatorname{Hom}(P, \ast)$ takes surjections to surjections.
Want to show: $Q$ is injective if and only if injections from $Q$ always split. The $\Longrightarrow$ direction is straightforward. The $\Longleftarrow$ direction, not so much. (Dummit and Foote relegates this to the exercises and uses the nontrivial fact that every module is contained in an injective module. I imagine this is to parallel their proof that $P$ is projective iff surjections to $P$ always split, which uses the fact that every module is a quotient of a free, hence projective, module.)
It bothered me that these dual concepts don't have "dual" proofs, so I came up with one using pushouts and pullbacks (D&F introduce these in exercise 27, if you have the book). The argument is rather straightforward, so I feel like something has to be wrong, but I can't figure out what it is. Here it is:
Suppose injections from $Q$ always split, suppose $i: A \to B$ injects, and suppose $f \in \operatorname{Hom}(A,Q)$. Consider the pushout of $i$ and $f$: $$M = B \oplus Q/\{(i(a), -f(a)): a \in A\}.$$ Then we get the following maps "for free": $j: B \to M$ and $g: Q \to M$, with $ji = gf$. (At this point, we haven't used any of our hypotheses.) Now, it's straightforward to check that $g$ injects because $i$ does, hence $g$ splits and we get a map $h: M \to Q$ with $hg = 1$. Thus, $hj$ lifts $f$.
Verification that $g$ injects because $i$ does: If $g(q) = 0$, then $(0,q) = (i(a), -f(a))$ for some $a$. But $i(a) = 0$ implies that $ a = 0$, and this in turn implies that $-f(a) = 0$, so that $q = 0$.
(A completely analogous proof using the pullback shows that: projections to $P$ always split $\implies P$ is projective.)
Where is the flaw in this proof?
Best Answer
Your proof is correct. I imagine Dummit and Foote do it their way because they don't want to have to discuss pushouts and pullbacks, and because the facts that every module is a quotient of a projective module and a subobject of an injective module are very important in their own right. But your proof has the nice advantage of applying in more general contexts where that proof would fail (more precisely, your proof can be made to work in any abelian category).