Why a projective module is a projective cover for its largest semisimple quotient? That is  why the projection on the quotient is an essential morphism in this case?

2$\begingroup$ This question is not quite ontopic on this site, as explained in the FAQ. The FAQ suggests a few other places where your question will be much more at ease. Good luck! $\endgroup$– Mariano SuárezÁlvarezOct 28, 2012 at 4:14

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1 Answer
Answer: if Q is a submodule of a projective module P which projects surjectively on the largest semisimple quotient of P, then Q projects surjectively on each simple quotient of P, and hence Q lies outside of any maximal submodule of P  contradiction.

2$\begingroup$ In other words, because the kernel is contained in the radical (because the map is surjective and the quotient has zero radical!) $\endgroup$ Oct 28, 2012 at 4:13