In “Cover of a direct summand” it is asked to show that if a direct sum has a projective cover, and if one of the summands has a projective cover, then so does the other. I gave a solution that works for any class $\mathcal{X}$ closed under direct sums and direct summands, however I use the smaller cover in an essential way.
Is it true that every summand of a module with a projective cover itself has a projective cover?
Suppose $\mathcal{X}$ is a class of modules. A morphism $f:X \to Y$ is called an $\mathcal{X}$-cover of $Y$ if (a) $X \in \mathcal{X}$, (b) whenever $g:X' \to Y$ is a morphism with $X' \in \mathcal{X}$ there is a morphism $h:X' \to X$ such that $X' \xrightarrow{h} X \xrightarrow{f} Y = X' \xrightarrow{g} Y$, and (c) if $h:X \to X$ is such that $X \xrightarrow{h} X \xrightarrow{f} Y = X \xrightarrow{f} Y$ then $h$ is an automorphism of $X$.
If $\mathcal{X}$ is closed under direct sums and direct summands and $Y \oplus Z$ has an $\mathcal{X}$-cover, must $Y$ have a $\mathcal{X}$ cover?
(If it did, even by additional hypothesis, then so would $Z$.)