Let $A \xrightarrow{\phi} B$ be an injective map of $R$ modules. Let $P$ be a projective $R$ module. Let $a \in A$ and $p \in P$. And suppose there is a bi-linear map $A \otimes_R P \to Q$ sending $a$ and $p$ to something nonzero, where $Q$ is some $R$ module.
Question: How can I prove that there is a bi-linear map from $B \otimes_R P$ sending $\phi(a)$ and $p$ to something nonzero, where $Q'$ is some other $R-$ module?
Useful Info/Motivation: When $P$ is a free $R$-module over $s_1,...,s_n$, I can pick $Q$ to be $A$ and $Q'$ to be $B$. In this case there is the map above on $A \otimes P$ sending $a \otimes (\sum r_i s_i)$ to zero if there is an $r_{i_0}$ such that $r_{i_0} a=0$. One can choose the map $B \otimes P \to B$ to be the map that sends $b \otimes (\sum r_i s_i)$ to $b r_{i_0}$.
My question above is an attempt to generalize this result enough that I can just use the universal property for projective modules. Please don't just use the fact that projective modules are a direct summand of a free module in defining your maps( you would just need to send the complement of the free part to 0). My aim is to directly use the universal property for projective modules in doing my question.
Well, a rather perverse proof without free modules would go as follows. $P$ being flat means that $L_1 (-\otimes_R P) (M) = 0$ for all $M$. Then we can invoke the "balancing of Tor" $L_1 (-\otimes_R P) (M) \cong L_1 (M\otimes_R -) (P)$, and finally the fact that $L_n F (P) = 0$ for $n > 0$ for any right exact $F$ and any $P$ projective (this follows from the construction of left derived functors via projective resolutions).
– Oct 19 '16 at 02:45