Is it true that in the category $R$-Mod of $R$-modules (and $R$-module homomorphisms), the diagram $$ M \longrightarrow M \otimes_{R} N \longleftarrow N,$$ where the arrows are the maps $m \to m \otimes 1$, $n \to n \otimes 1$, is a coproduct diagram?
I know that for commutative rings instead of $R$-modules this is true, but I don't why this is different. A detailed explanation for a noob in category theory would be nice. (not hw, I would just like to get this straight, thank you very much).
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$\begingroup$There is no way of defining the inclusion maps (what is $1 \in M$?). When $M,N$ are free of ranks $n,m$, we have that also $M \otimes_R N$ is free of rank $n \cdot m$. But the direct sum i.e. coproduct $M \oplus N$ is free of rank $n+m$. So they are not isomorphic when $n,m>0$. In fact, there is a big difference between them.
There is the following connection between tensor products and direct sums (see ): For every $m \in M$ let $i_m : N \to \oplus_{m \in M} N$ denote the inclusion. Similarly, for every $n \in N$ let $j_n : M \to \oplus_{n \in N} M$ denote the inclusion. Then $$M \otimes_R N = \bigoplus_{m \in M} N \oplus \bigoplus_{n \in N} M / \langle i_m(n) - j_n(m) \rangle_{m \in M, n \in N}.$$
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