A difficulty in understanding the proof of distributivity of tensor products over direct sums for modules. Here is the proof:


But I do not understand the following:
1-why the function needed to be bilinear to use the universal property?
2- what is he doing starting from the paragraph that starts with the statement "In the other direction ...." ?
 A: *

*In order to define maps from the tensor product of $M$ and $N$, one needs a bi-linear map $\phi:M \times N \rightarrow G$.


This is how one commonly defines a tensor product of modules via a universal property. Since $M \otimes N$ is a quotient group of $M \times N$, one needs to make sure a map from $M \otimes N \rightarrow G$ is well-defined, which is the case if there is a bilinear map $\phi:M \times N \rightarrow G$. (Wikipedia refers to the property of $\phi$ as "well-balanced").


*The author wants to first find a map $\phi:(M\otimes M')\oplus N \rightarrow (M\otimes N) \oplus (M' \otimes N)$, and then a map $\phi^{*}:(M\otimes N) \oplus (M' \otimes N) \rightarrow (M\otimes M')\oplus N$. By defining these maps, and showing that they are inverses of each other, this makes them isomorphisms.

A: The universal property of tensor product basically says that there is a one-to-one correspondence between
$$\matrix{M\times N\to T&R\text{-bilinear}\\ \hline M\otimes N\to T& R\text{-linear}} $$
maps where $M, N, T$ are arbitrary $R$-modules.
That's why for finding $(M\oplus M')\otimes N\to (M\otimes N)\oplus (M'\otimes N)$ we actually only have to define a bilinear map $(M\oplus M')\times N\to(M\otimes N)\oplus(M'\otimes N)$. 
Similarly, for the other direction, the two bilinear maps $M\times N\to(M\oplus M')\otimes N$ and $M'\times N\to (M\oplus M') \otimes N$ induce $R$-homomorphisms with domains $M\otimes N$ and $M'\otimes N$, which together induce a map from their direct sum, actually by the universal property of coproducts (and that one indeed doesn't involve bilinear maps). 
