# Weibel, Step in the proof of proposition 3.2.9

I don't understand a step of a proof in Weibel's book. Proposition 3.2.9 in Weibel's homological algebra book states that

Assume $$T$$ is a flat $$R$$-algebra. Then for all $$T$$-modules $$C$$ and $$R$$-modules $$A$$ it holds that \begin{align*} Tor^R_n(A,C) = Tor_n^T(A\otimes_R T,C)\,. \end{align*}

The proof goes as follows. Take a projective resolution $$P_*$$ of $$A$$. Since $$T$$ is flat $$P_*\otimes_R T$$ is a resolution of $$A \otimes_R T$$. Next Weibel states that each $$P_n \otimes_R T$$ is a projective $$T$$-module. I don't see why this is the case. Question: Why are the $$P_n\otimes_RT$$ projective?

My thoughts: I know that left adjoints preserve projective objects when their right adjoint is exact. But to use this I would need that $$Hom(T,-)$$ is exact, i.e. that $$T$$ is projective, and flat modules are not projective in general. I think the statement can be saved by using that $$Tor$$ can be computed by using flat resolutions. If $$f$$ is monic, then so is $$(P_n \otimes_R T)\otimes_T f$$ since tensor product is associative and $$P_n$$ is flat.

You need to pay attention to the category you are working in. $$T$$ is certainly a projective $$T$$-module, and $$\textrm{Hom}_T (T, -)$$ is exact as a functor from $$T$$-modules to $$R$$-modules, so $${-} \otimes_R T$$ sends projective $$R$$-modules to projective $$T$$-modules.