I am taking a course on commutative algebra, and we just defined the Tor functor using projective resolutions of a module. The definition we have is:
Let $R$ be a commutative ring with unity. Let $M$ be an $R$-module and $\varepsilon:P_*\rightarrow M$, $\eta:Q_*\rightarrow M$, two projective resolutions of $M$. For all $R$-modules $N$, we may define the chain complexes $P_*\otimes N \rightarrow M\otimes N \rightarrow 0$ and $Q_*\otimes N \rightarrow M\otimes N \rightarrow 0$. We then define the functor Tor$_i^R(M,N)=H_i(P_*\otimes N)$, where $H_i(P_*\otimes N)$ denotes the i-th homology group of the chain complex $P_*\otimes N \rightarrow M\otimes N \rightarrow 0$. We have $H_i(P_*\otimes N)\cong H_i(Q_*\otimes N)$.
The claim is that this definition is independent of the projective resolution chosen. The proof is left as an exercise for us. I was able to show $H_0(P_*\otimes N)\cong M\otimes N\cong H_0(Q_*\otimes N)$, but couldn't show the rest. The hint is to use the horseshoe lemma to somehow construct an isomorphism between the homology groups, but I don't know where to start exactly. I would very much appreciate it, if someone could give me hints or lead me in the right direction. Thanks!