Can we say that the following diagram is commutative?

Question

Let $R$, $T$ be two commutative rings, $f :R\to T $ a morphism of rings, $M$ be an $R$-module, $I$ be an ideal of $T$ and $i :I\to T $ be the canonical injection.

We know that

$$ (M\otimes_{R} T)\otimes_{T} I\cong M\otimes_{R}I $$ and $$ (M\otimes_{R} T)\otimes_{T} T\cong M\otimes_{R}T $$ Can we say that the following diagram is commutative?

$$\require{AMScd} \begin{CD} (M\otimes_{R} T)\otimes_{T} I @>1_{M\otimes_{R}T}\otimes f>> (M\otimes_{R} T)\otimes_{T} T \\ @VV{\cong}V @VV{\cong}V \\ M\otimes_{R}I @>1_{M}\otimes f>> M\otimes_{R}T \end{CD} $$

Answer 1

The isomorphism on the left does $x\otimes t\otimes u\mapsto x\otimes tu$; the isomorphism on the right does $x\otimes t\otimes u\mapsto x\otimes tu$.

The top map does $x\otimes t\otimes u\mapsto x\otimes t\otimes u$; the bottom map does $x\otimes t\mapsto x\otimes t$.

So yes, the diagram is commutative.

Answer 2

Formally, this is a case of the question whether the functors $M\otimes_R (-)$ and $M\otimes_R T\otimes_T (-)$ of $T$-modules are naturally isomorphic. And indeed they are, being constructed by composing $M\otimes_R(-)$ with, respectively, the identity functor and the functor $T\otimes_T (-)$. But the latter two are well known to be naturally isomorphic; a bilinear map out of $T\otimes N$ is naturally identified with a linear map out of $N$.