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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!!

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    $\begingroup$ Because the zero-th derived functor of a right-exact functor is the original functor? $\endgroup$ – Angina Seng Sep 4 '19 at 18:36
  • $\begingroup$ 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. $\endgroup$ – Ayman Hourieh Sep 4 '19 at 19:20
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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.

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