# Can one prove the existence of tensor product without explicitly constructing it? [duplicate]

R is a ring with 1.

We construct tensor product $M \otimes N$ of right R-module $M$ and left R-module $N$ to basically be able to state its universal property that any R-bilinear map from $M\times N$ to R-module P factors through an R-linear map from $M \otimes N$ to P. Atiyah and McDonald even mention that one can forget the construction if one wants to, we only need the universal property.

Given all this, can one do away with the process of constructing $M \otimes N$ and rather just prove the existence of one.

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## marked as duplicate by Grigory M, USER91500, user1729, Najib Idrissi, user88595May 27 '14 at 9:08

This question was marked as an exact duplicate of an existing question.

– Grigory M May 25 '14 at 20:40
Yes, the linked question gives a complete answer. I would vote to close if there weren't a bounty. – Kevin Carlson May 25 '14 at 22:18

You can do it using category theory. Consider the functor $F:Ab\to Set$ given by sending $$A\to Bil_{R}(M,N,A)$$ where the notation on the right stands for bilinear maps into $A$. The tensor product is a representing object for $F$. That is $$Hom(M\otimes_{R}N,A)\simeq F(A)$$ where the bijection is natural. I suspect the existence of such an object follows from Brown representability, but I'll let you check the details for yourself.