It is known that $\operatorname{Tor}$ is independent of the choice of the resolution.
More specifically, I am trying to do the exercise 1 (c) of Vick's homology theory. The author gives the definition of free resolution right after proving the right exactness of tensor.
That means I need a direct proof of it without a lengthily reference of commutative algebra (which I poorly known).
(I copy that page in the book below for the definition and properties.)
It is helpful if anyone could also give me some hints about Exercise 1 (b) and 2.
For 1 b), the exact sequence $0 \to A \to A \to 0$ is a free resolution of $A$, tensoring by $B$ and taking homology gives you $Tor(A,B)=0$.
For 2, use the fact that $\mathbb{Z}$ is a commutative ring, that gives you an isomorphism of abelian groups $A \otimes_{\mathbb{Z}} B = B \otimes_{\mathbb{Z}} A$, for all abelian groups $A$ and $B$.
For 1 c), use the fact that if $0 \to C \to B \to A \to 0$ and $0 \to C' \to B' \to A \to 0$ are free resolutions, you can lift the identity $1_A:A \to A$ to a morphism $f:B \to B'$ such that the corresponding diagram commutes, and you may find a morphism $g: C \to C'$ that makes the new diagram commute. For there it is easy to find the required isomorphism.
I also recommend you to take a look at commutative algebra basics in order for you to understand better the homological algebra basics.