Ok, today we are interested in finding a projective cover of a given ${R}$-module, which can be done under certain circumstances. (Injective hulls, by contrast, always exist.) The setting in which we are primarily interested is the case of ${R=k[G]}$ for ${k}$ a field. If the characteristic ${k}$ doesn’t divide ${|G|}$, then ${R}$ is semisimple and every module is projective, so this is trivial. But in modular representation theory one does not make that hypothesis.  Then taking projective envelopes of simple objects gives the indecomposable projective objects.

Projective Covers

So, fix an abelian category ${\mathcal{A}}$ that has enough projectives (i.e. for ${A \in \mathcal{A}}$ there is a projective object ${P}$ and an epimorphism ${P \rightarrow A}$) where each object has finite length.  Example: the category of finitely generated modules over an artinian ring.

An epimorphism ${f:A \rightarrow B}$ is called essential if for each proper subobject ${A' \subset A}$, ${f(A') \neq B}$. A projective cover of ${A}$ is a projective ${P}$ with an essential map ${P \rightarrow A}$

Theorem 1 Each object in ${\mathcal{A}}$ has a projective cover.

Pick ${A \in \mathcal{A}}$ and write ${A}$ as a quotient of some projective ${L}$, via a map ${L \rightarrow A}$. Consider the collection of subobjects ${R \subset L}$ such that ${im(R) = A}$; since ${L}$ has finite length, we can choose a minimal element ${F}$.

We thus have an epimorphism ${L \rightarrow L/F \rightarrow A}$. I claim that ${L/F}$ is the projective cover. To see this, choose a subobject ${G \subset L}$ such that the map ${G \rightarrow L/F}$ is epi, and ${G}$ is the minimal such subobject. There is a commutative diagram Since ${L}$ is projective, we can find a lifting ${L \rightarrow G}$ making the following diagram commutative: So if we take the kernel ${K}$ of ${L \rightarrow G}$, we have a commutative diagram (note that ${L \rightarrow G}$ is epi by minimality of ${G}$): By commutativity, we have ${F \supset K}$. Since ${G \simeq L/K}$ and ${G \rightarrow L/F}$ is essential, so is ${L/K \rightarrow L/F}$ and consequently ${L/K \rightarrow L/F \rightarrow A}$ essential. This implies ${K=F}$ by the above assumptions, so ${G \simeq L/F}$ in the above commutative diagram. It now follows that since we have the monomorphism ${G \rightarrow L}$ of which ${L \rightarrow L/F \rightarrow G}$ is a retraction, that ${G}$ is a direct summand of ${L}$ and consequently projective. We’re consequently done. There is also uniqueness:

Proposition 2 The projective cover is unique up to isomorphism.

Given two projective covers ${P, P'}$ of ${A}$ with essential maps ${P \rightarrow A}$, ${P' \rightarrow A}$, we can do a lifting both ways to get a commutative diagram. By the essentiality, the map ${P \rightarrow P'}$ is epi and consequently by projectivity, split. But the extra direct factor in ${P}$ aside from ${P'}$ means that ${P \rightarrow A}$ isn’t essential after all, contradiction.

Uniqueness works in any abelian category by the above proof.  Source: Jean-Pierre Serre, Linear Representations of Finite Groups, Part III.

By the way, xymatrix is definitely a more versatile package for commutative diagrams than amscd. See, e.g., James Milne’s manual.