Tensor product over several modules

Question

Let $M,N,P$ be both $R$ modules, where $R$ is a commutative ring with $1$.
How could one show $$M\otimes N\otimes P\cong (M\otimes N)\otimes P$$ where the left hand side is defined by trilinear maps.
What I am thinking is to construct a linear map and its inverse. Define: $\varphi: M\otimes N\otimes P\rightarrow (M\otimes N)\otimes P:m\otimes n\otimes p\mapsto (m\otimes n)\otimes p$. This is well defined by the universal property.
What seems natural to me is to define its inverse as $\phi: (m\otimes n)\otimes p\mapsto m\otimes n\otimes p$. But how could one show the well definedness of this map?

Answer 1

For each $p \in P$, the map \begin{align} M \times N &\to M \otimes N \otimes P \\ (m,n) & \mapsto m \otimes n \otimes p \end{align} is bilinear, and so there is a unique linear map $$ f_p \colon M \otimes N \to M \otimes N \otimes P $$ such that $f_p(m \otimes n) = m \otimes n \otimes p$ for all $m \in M$ and $n \in N$.

The uniqueness allows us to prove that $f_{rp_1+p_2} = rf_{p_1}+f_{p_2}$ for all $p_1,p_2 \in P$ and $r \in R$.

Hence, the map \begin{align} (M \otimes N) \times P &\to M \otimes N \otimes P \\ (t,p) & \mapsto f_p(t) \end{align} is bilinear, so that there is a linear map $$ \phi \colon (M \otimes N) \otimes P \to M \otimes N \otimes P $$ such that $\phi(t \otimes p) = f_p(t)$ for all $t \in M \otimes N$ and $p \in P$. In particular: $$ \phi((m \otimes n) \otimes p) = f_p(m \otimes n) = m \otimes n \otimes p. $$