# If $M$ is indecomposable projective, then $\operatorname{rad}\hom(-,M)=\hom(-,\operatorname{rad}M)$

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?