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Let $R$ be a commutative ring and $f\colon A\rightarrow B$, $f'\colon A'\rightarrow B'$ two maps of $R$-modules.

Is there a way to express the cokernel $\operatorname{coker} (f\otimes f')$ in terms of the cokernels of $f$ and $f'$?

By tensoring $f$ with the identity on $R$, we get $\operatorname{coker}(f)\cong \operatorname{coker}(f\otimes \operatorname{id} )$, so in particular we do not have $\operatorname{coker}(f\otimes \operatorname{id} )=\operatorname{coker}(f)\otimes \operatorname{coker}(\operatorname{id} )$ as $\operatorname{coker}(id)=0$.

Davide Giraudo
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Kathrin L.
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    Why the downvote? What makes this (http://math.stackexchange.com/questions/145819/is-the-image-of-a-tensor-product-equal-to-the-tensor-product-of-the-images) more worth asking than mine? – Kathrin L. Jan 27 '17 at 12:23
  • Check out Bourbaki, A II.3.6 Proposition 6. – Fred Rohrer Feb 03 '17 at 08:15

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