Superfluous condition on universal property of tensor product Let $R$ be a ring and $M$ be a right $R$-module and $N$ be a left $R$-module. Let $\phi:M\times N \rightarrow M\otimes_R N$ be the canonical middle linear map. Here is a theorem in my text under the above setting: > Dummit&Foote p.366 > > Let $D$ be an abelian group and $\iota : M\times N \rightarrow D$ be a middle linear map. > > Assume that for any abelian group $A$ and a middle linear map $f:M\times N\rightarrow A$, there exists a unique group homomorphism $\Phi:D\rightarrow A$ such that $\Phi\circ \iota = f$. > > Assume also that **the image of $\iota$ generates $D$**. > > Then, there exists a unique group isomorphism $f:M\otimes_R N\rightarrow D$ such that $\iota=f\circ \phi$. But isn't the second condition superfluous? Exactly where the condition "the image of $\iota$ generates $D$" is used to prove the universal property? Isn't the tensor product just a coproduct?

The condition is not needed. Without the condition, $D$ has the same universal property as the tensor product, hence they are canonically isomorphic.