Why is $Tor_0(A,B) = A\otimes B$?

Question

We define a free resolution of an abelian group $A$ to be a level-wise free chain complex $\tilde A$ such that $H_0 (\tilde A) = A$.

Also, we define $Tor_k (A,B)$ to be $H_k(...\rightarrow \tilde{A_2}\otimes B \rightarrow \tilde{A_1}\otimes B \rightarrow \tilde{A_0}\otimes B \rightarrow 0 \rightarrow ...)$ for some free resolution $\tilde{A}$ of $A$.

I saw in multiple places that $Tor_0(A,B) = A\otimes B$, but I couldn't quite formalize why, and all proofs I found used abstractions I don't quite grasp.

Can anyone please give a proof of this fact?

Answer 1

By definition, $\operatorname{Tor}_0(A,B)$ is the cokernel of the map $\tilde A_1\otimes B\to \tilde A_0\otimes B$. Since tensoring with $B$ commutes with cokernels, this is precisely $\operatorname{coker}(\tilde A_1\to\tilde A_0)\otimes B$. Now observe that $A=H^0(\tilde A)=\operatorname{coker}(\tilde A_1\to\tilde A_0)$ holds and you are done.