Let $R$ be a commutative ring with identity, and $M$ an $R$-module. Then $$R\otimes_R M \simeq M$$ This can be easily shown directly by construction of the isomorphism, namely $r\otimes m \mapsto rm$ for $r \in R$ and $m \in M$. Indeed, let $rm=0$. Then $r \otimes m = r\left(1\otimes m\right) = 1 \otimes rm = 1 \otimes 0 = 0$. Hence the homomorphism is injective (I skip the proof that the map in question is indeed a homomorphism; this is straight forward). On the other hand, for any $m\in M$, $1\otimes m$ gets mapped to $m$, thus showing surjectivity.
My question is, does anyone know a way to prove the same result without explicit construction of the isomorphism, i.e. via the universal property?