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Let $k$ be an algebraically closed field and $A$ be a finite dimensional $k$-algebra. If $M$ is finitely generated $A$-module, the contravariant hom functor $\hom_A(-,M)$ is an element of the functor category $\operatorname{Fun}:=\operatorname{Ab}^{(\operatorname{mod}_A)^\text{op}}$. Denote by $\operatorname{rad}\hom_A(-,M)$ the intersection of all maximal subfunctors of $\hom_A(-,M)$. If $M$ is indecomposable, this has a nice description: If $N=\bigoplus N_i$, then $\operatorname{rad}\hom_A(N,M)$ consists of all morphisms $f$ that have no section, i.e., there is no morphisms $s\colon M\to N$ such that $fs=\operatorname{id}_M$. Denote by $\operatorname{rad} M$ the intersection of all maximal submodules of $M$.

Claim: If $M$ is indecomposable projective, then $\operatorname{rad}\hom_A(-,M)=\hom_A(-,\operatorname{rad} M)$.

Proof: $\supseteq$) This direction is easy: If $f\colon N\to M$ has a section $s$, then $f$ must be surjective, so in particular $M=f(N)\nsubseteq\operatorname{rad}M$, and $f$ cannot be in $\hom_A(-,\operatorname{rad} M)$.

For the other direction I have no idea. I should be using projectiveness somewhere, but where?

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