Let $(F_{i})_{i\in I}$ be a family of left R-modules and E is a right R-module.
(a) Define a homomorphism $\pi :E\otimes_{R}(\prod_{i\in I} F_{i})\to \prod_{i\in I}(E\otimes_{R} F_{i})$
question: Show that if E is a free right R-module, then $\pi$ is injective.
My thoughts: set $\pi(e\otimes(\prod_{i\in I}f_{i}))=\prod_{i\in I}(e\otimes_{R}f_{i})$, I think this definition is reasonable,but I still have no idea how to prove it is injective. Can someone help me solve this problem?
As summarized in the comment, to prove injectivity, one only need to show that it is true for $R$.
Now, as map as defined, for $r\in R$, one has $ r\otimes(\prod_{i\in I}f_{i})=1\otimes r(\prod_{i\in I}f_{i})=1\otimes(\prod_{i\in I}rf_{i})=0$ if and only if $\prod_{i\in I}rf_{i}=0$, if and only if $rf_i=0$ for all $i$, if and only if $1\otimes rf_i=0$ for all $i$, if and only if $\prod_ir\otimes f_i=0$