# Well definedness of a map in tensor product.

Suppose $M,N$ and $P$ be $R$-modules. For each fixed $p \in P$ construct a map $\varphi : M \otimes_R N \longrightarrow (M \otimes_R N) \otimes_R P$ defined by $\varphi (m \otimes n) = m \otimes (n \otimes p),\ m \in M,\ n\in N$.

Now my question is " Is the above map well-defined?" For doing this we take $m,m' \in M$ and $n,n' \in N$ with $m \otimes n = m' \otimes n'$. In order to prove the well-definedness we have to show that $\varphi (m \otimes n)= \varphi (m' \otimes n')$ i.e. $m \otimes (n \otimes p) = m' \otimes (n' \otimes p)$. Right?

But how can I show that? I know that $M \otimes_R N$ is $F/G$ together with a bilinear map $\theta : M \times N \longrightarrow F/G$ defined by $\theta(m,n) := m \otimes n = (m,n) + G$, where $F$ is a free $R$-module on $M \times N$ and $G$ is a submodule of $F$ generated by all the elements of the following type $:$

$(i)$ $(m+m',n)-(m,n)-(m',n),\ m,m' \in M,\ n \in N$,

$(ii)$ $(m,n+n')-(m,n)-(m,n'),\ m \in M,\ n,n' \in N$,

$(iii)$ $(am,n) - a(m,n), m \in M,\ n\in N,\ a \in R$,

$(iv)$ $(m,an) - a(m,n), m \in M,\ n\in N,\ a \in R$.

From this large structure of $M \otimes N$ how can I deduce the well-definedness of the above map?

I really need help in this regard as I have confused with such a big structure and can't manage to find appropiate reason behind it. Please help me.

• Use the universal property of the tensor product. All you need to do is to show that $M\times N\to (M\otimes N)\otimes P$ defined by $(m,n)\to (m\otimes n)\otimes p$ is bilinear. – Lord Shark the Unknown Apr 27 '18 at 8:30
• It's a theorem that if $\phi:M\times N\to X$ is $R$-bilinear, then there is a unique $R$-module homomorphism $\psi:M\otimes N\to X$ with $\psi(m\otimes n)=\phi(m,n)$. This is the most basic theorem in the study of tensor products. – Lord Shark the Unknown Apr 27 '18 at 8:35