$Hom_{A-Alg}(Sym(M),B)\cong Hom_{A-mod}(M,B)$

Question

Let A be an ring. M be an A-module. Let $Sym (M)$ be the Symmetric algebra of M over A. Let B be an A-algebra.

Why is $Hom_{A-Alg}(Sym(M),B)\cong Hom_{A-mod}(M,B)$

One way is clear - If we have a map $\phi:Sym(M)\rightarrow B$, we get a map $M\rightarrow B$ by composing $\phi$ with the the map $i:M\rightarrow Sym(M)$.

How do we get the other way correspondence? I.e., if we have a A-linear map $f:M\rightarrow B$. how do we get a map from $Sym(M)\rightarrow B$

Answer 1

I was just stupid to ask the question. Just take the map $m\otimes m \otimes m ... \otimes m $ to $f(m).f(m)...f(m)$ and extend it linearly.