Cover of a direct summand Let $L,N$ be $R$-modules. If $L$ and $L \oplus N$ have projective covers, is it true that $N$ admits a projective cover?
 A: I'm not sure how much you've studied projective covers. I use the universal property definition of cover restricted to the class of projective modules, but it is a (slightly long) exercise to prove all of my claims using the “superfluous kernel” definition.
Let $f:P \to L \oplus N$ be a projective cover of $L \oplus N$ and let $g:Q \to L$ be a projective cover of $L$.
Since $g$ is a projective cover, the map $P \to L \oplus N \to L$ must factor through $g$ via a split epimorphism, so there is some $h:P \to Q$ with $P \xrightarrow{h} Q \xrightarrow{g} L = P \xrightarrow{f} L \oplus N \to L$ and $0 \to \ker(h) \to P \xrightarrow{h} Q \to 0$ is split exact.
We can extend this to a map $\ker(h) \hookrightarrow P \xrightarrow{f} L \oplus N \to N = \ker(h) \xrightarrow{k} N$.
Since $\ker(f) = \ker(k) \oplus \ker(g)$ we get that $\ker(h)$ is superfluous in $\ker(k)$ so that $k$ is a projcetive cover of $N$.
Question: Suppose only that $L\oplus N$ has a projective cover. Can we conclude that $L$ and $N$ have projective covers? I use the smaller projective cover in an essential way to split $P$ correctly.
