# Why does $\mathrm{Tor}_0^R(M,N)\cong M\otimes_R N$

I was reading this question Why does $\mathrm{Tor}_0^R(M,N)\cong M\otimes_R N$? While I understand most of the steps here, I'm not quite sure why we can say that $$\text{im} (P_1\otimes N\to P_0\otimes N) =\alpha_1(P_1)\otimes N$$ If I recall correctly, the image of a tensor product is not necessarily equal to the tensor product of the images, so I see no reason why this should hold in general.

I'm sure I'm missing something very simple, thanks for any help!!

• Because the zero-th derived functor of a right-exact functor is the original functor? – Angina Seng Sep 4 '19 at 18:36
• It's likely that someone learning about $\operatorname{Tor}$ for the first time doesn't know what a derived functor is. Either way, see my answer below. I use slightly different notation. Hopefully this clears things up. – Ayman Hourieh Sep 4 '19 at 19:20

Start with a projective resolution $$\cdots \to P_2 \to P_1 \to M \to 0.$$
Tensor with $$N$$ and remember that $$\_ \otimes_R N$$ is right exact. This gives the following exact sequence $$P_2 \otimes_R N \xrightarrow{\varphi_2} P_1 \otimes_R N \xrightarrow{\varphi_1} M \otimes_R N \to 0$$
By exactness, $$\operatorname{im} \varphi_2 = \ker \varphi_1$$. Thus, $$\operatorname{Tor}_0^R(M, N) = (P_1 \otimes_R N) / \ker \varphi_1$$, and this isomorphic to $$M \otimes_R N$$ by the first isomorphism theorem.