Let $A$ be a commutative ring with identity, and let $(M_i)_{i \in I}$ be a family of $A$-modules (for some index set $I$). I am trying to show that for all $A$-modules $N$,
$N \otimes (\bigoplus_{i \in I} M_i) \cong \bigoplus_{i \in I}(N \otimes M_i)$,
i.e. the tensor product distributes over the direct sum. So far I have shown that the map $f : N \times (\bigoplus_{i\in I} M_i) \to \bigoplus_{i\in I}(N \otimes M_i)$ given by $f(n, (m_i)_{i\in I}) = (n \otimes m_i)_{i\in I}$ is $A$-bilinear, and hence by the universal property of the tensor product, we induce a module homomorphism $\hat{f} : N \otimes (\bigoplus_{i\in I} M_i) \to \bigoplus_{i\in I}(N \otimes M_i)$ which furthermore satisfies $\hat{f}(n \otimes (m_i)_{i \in I}) = (n \otimes m_i)_{i\in I}$ for all $n \in N$ and all $(m_i) \in \bigoplus_{i\in I} M_i$.
It seems intuitive that $\hat{f}$ is the desired isomorphism, but I can't seem to argue the bijectivity. I'm not used to using tensor products, but I know that that the elements of the tensor product aren't all simply of the form $n \otimes (m_i)_{i \in I}$, so even injectivity is difficult for me to see. How can I show $\hat{f}$ is bijective?