$\Lambda$ is an artin algebra, $M$ a finitely generated $\Lambda$-module and $P$ a projective indecomposable $\Lambda$-module. Show: $$\text{Hom}_\Lambda(P,M) \neq 0 \iff P/\underline{r}P \text{ is a composition factor of $M$}$$ where $\underline{r} = \operatorname{rad}\Lambda$.
$\Leftarrow$:
Since $P/\underline{r}P$ is a composition factor of $M$, then we have the following for some composition series of $M$: $$M = M_0\supseteq M_1 \supseteq M_2 \supseteq ... \supseteq P \supseteq \underline{r}P \supseteq ... \supseteq M_{n-1} \supseteq M_n = (0)$$ Then let $\phi : P \rightarrow M$ be the inclusion homomorphism, which is clearly not zero. So then $\text{Hom}_\Lambda(P,M) \neq 0$
$\Rightarrow$:
This direction is giving me a headache. I know that $P \rightarrow P/\underline{r}P$ is a projective cover of $\underline{r}P$ and thought I might use that, but I cannot see how. Any hints or complete answers are greatly appreciated as I'm having my exam tomorrow and cannot figure this out.
$\Leftarrow:$ I think there is a place not right in your words: $P/rP $ is a composition factor of $M$ means that: for some composition series $0=M_n \subseteq M_{n-1} \subseteq \cdots \subseteq M_1 \subseteq M_0=M$ of $M$, there is an $i$ such that $M_i/M_{i+1} \cong P/rP$. So you can define a surjective map $P \rightarrow M_i$, then $\mathrm{Hom}_{\Lambda}(P,M) \not =0$.
$\Rightarrow:$ There is a conclusion here: Let $M,N$ be two modules such that $M$ has a simple top $S(M)$, given a nonzero map $f: M \rightarrow N$, then $S(M)$ must occur in a composition series of $N$. (Now $P$ is an idecomposable projective module, so it has a simple top $P/rP$, also $\mathrm{Hom}_{\Lambda}(P,M) \not=0$, so you can get the result.) For the proof of the conclusion, you can see here: The map between modules having unique top and socle.