Projective objects in abelian categories having non-trivial morphisms I'm trying to read an article about $p$-adic groups (based on the lectures of Joseph Bernstein), and I'm struggling to understand a certain argument regarding projectives objects in abelian categories wich goes like this:
let $X$ and $Y$ be two objects in an abelian category $M$, such that they are both projective objects and $X$ is a projective generator. Then $\hom(Y,X)$ is non-trivial.
If you could provide an explanation with as little categorical jargon as possible I would be grateful as my knowledge regarding category theory is quite limited.
 A: Presumably you want to assume that $Y$ is nonzero.
But even then, this is not true in a general abelian category.
For example, take the opposite category of the category of abelian groups. Take $X=\mathbb{Q}/\mathbb{Z}$ and $Y=\mathbb{Q}$. Both are injective in the category of abelian groups, and so projective in the opposite category. And $X$ is an injective cogenerator in the category of abelian groups, and so a projective generator in the opposite category.
But for $\operatorname{Hom}(Y,X)$ to be nontrivial in the opposite category, we need $\operatorname{Hom}(X,Y)$ to be nontrivial in the category of abelian groups. But there are no nontrivial group homomorphisms $\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}$, since $\mathbb{Q}/\mathbb{Z}$ is a torsion group, and $\mathbb{Q}$ is torsion-free.
Maybe you are missing some assumption on the abelian category?
A: A projective object is something which satisfies the following lifting property: $P$ is projective if for any surjection $A \to B$ and any morphism $P \to B$, there exists a morphism $P \to A$ making the triangle commute: an arrow $P \to A$ always exists in a diagram
\begin{align*}
&\ P \\
&\downarrow \\
A \twoheadrightarrow \ &B  
\end{align*}
Now a projective generator is a projective object such that any other object is a quotient of a direct sum of copies of it (e.g. $\mathbb Z$ is a projective generator in the category of finitely generated abelian groups, by the structure theorem). So we can write $Y = X^n/K$ for some object $K$ and natural number $n$. In other words, there is a short exact sequence
$$
0 \to K \to X^n \to Y \to 0
$$
for some object $K$.
Since $Y$ is projective, I get a morphism $Y \to X^n$, using
\begin{align*}
&Y \\
&\big{|}\big{|} \\
X^n \twoheadrightarrow \ &Y  
\end{align*}
If the morphism $Y \to X^n$ is trivial (i.e. zero), it means that $\text{id}_Y$ is trivial, i.e. $Y = 0$. If we exclude this case, we see that $Y \to X^n$ is non-trivial, which means that one of the compositions $Y \to X^n \xrightarrow{\pi_i} X$ with the projection to the $i^{\text{th}}$ factor must be non-trivial - this is because a morphism $Y \to X^n$ is uniquely determined by these compositions, and the zero morphism $Y \to X^n$ would then the same compositions with all the projections. So we conclude that $\text{Hom}(Y,X) \neq 0$.
Note that the zero object is projective, so one really should take care that $Y \neq 0$ in your set-up.
