# Why is this homomorphism on an infinite product well defined?

Browsing over some questions, I found that the natural homomorphism from $(\prod M_i)\otimes N\to \prod(M_i\otimes N)$ is given by $(\prod m_i)\otimes n\mapsto \prod(m_i\otimes n)$.

This of course seems very natural, but how does one know it is in fact well defined? The infinite product of representative from $M$ is giving me a difficult time accepting this, although I don't doubt it to be true. Thanks.

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Homomorphisms can't browse. –  joriki Mar 11 '12 at 7:50
The notation $\prod m_i$ is misleading: no multiplication is happening. –  Zhen Lin Mar 11 '12 at 9:11

Okay, in this answer I will assume you know the universal property satisfied by products, and that the category of $A$ modules has arbitrary products.

Note that for all $i$, you have a natural bilinear map $\prod({M_i}) \times N \rightarrow M_i \otimes N$. Thus, by the universal property of the tensor product $(\prod{M_i}) \otimes N$, for all $i$, you get a unique $A$-linear map $(\prod{M_i})\otimes N \rightarrow M_i \otimes N$. Then, by the universal property of products (since arbitrary products exist in the category of $A$-modules), you get an $A$-linear map $(\prod{M_i}) \otimes N \rightarrow \prod({M_i}\otimes N)$. This is your desired map.

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Oh, and my $A$ denotes a ring (the base ring). –  Rankeya Mar 11 '12 at 8:17
Dear Rankeya, the map in your third line is linear and definitely not multilinear so that the map at the beginning of the 5th line is not well-defined. The correct way to proceed is to first consider the bilinear maps $(\prod{M_i}) \times N \rightarrow M_i \otimes N$, to deduce linear maps $(\prod{M_i}) \otimes N \rightarrow M_i \otimes N$ and finally get the required map $(\prod{M_i}) \otimes N \rightarrow \prod (M_i \otimes N)$. –  Georges Elencwajg Mar 11 '12 at 10:51
Oops. I meant what you said, but wrote it wrong. Thanks for pointing it out. –  Rankeya Mar 11 '12 at 13:48

The projection maps $\pi_j: \prod_i M_i \rightarrow M_j$ induces a map $\pi_j \otimes id_N: (\prod_i M_i)\otimes N \rightarrow M_j\otimes N$ for each j. Now, the desired map follows from the universal property of the product $\prod_i (M_i\otimes N)$

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Ahh you just beat me to this answer... –  Rankeya Mar 11 '12 at 8:14