Superfluous condition on universal property of tensor product

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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?

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