Tor(A, B) is well-defined

Question

Let $A, B$ be abelian groups, and let \begin{align*} 0 \rightarrow C \xrightarrow i D \xrightarrow \pi A \rightarrow 0 \end{align*} be a free resolution of $A$ (i.e. a short exact sequence with $C$ and $D$ both free abelian). Then we have an exact sequence \begin{align*} C\otimes B \xrightarrow{i\otimes \text{id}} D\otimes B \xrightarrow{\pi\otimes \text{id}} A\otimes B \rightarrow 0. \end{align*} On page 70 of "Homology Theory," Vick defines $\text{Tor}(A, B)=\ker(i \otimes \text{id})$. Why is this independent of our choice of free resolution? This fact is given as an exercise, so it should be provable from this definition.

Answer 1

Here is a standard result in homological algebra, specialized to the case of abelian groups: suppose that $0 \to P_{1} \to P_{0} \to A \to 0$ and $0 \to Q_{1} \to Q_{0} \to C \to 0$ are chain complexes of abelian groups with each $P_{i}$ free and with the second complex exact. Then (a) any map $A \to C$ induces a chain map between these chain complexes: $$ \begin{array}{ccccccccc} 0 & \to & P_{1} & \to & P_{0} & \to & A & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & Q_{1} & \to & Q_{0} & \to & C & \to & 0 \end{array} $$ and (b) any two such chain maps are chain homotopic.

This is not hard to prove: choose a basis for the free abelian group $P_{0}$. For any basis element $x$, see where it goes under the composite $P_{0} \to A \to C$. Since the map $Q_{0} \to C$ is surjective (by exactness of the second sequence), there is an element $y$ of $Q_{0}$ with the same image as $x$, so send $x$ to $y$. Doing this for each basis element of $P_{0}$ defines a map $P_{0} \to Q_{0}$. Repeat this argument, essentially, to get a map $P_{1} \to Q_{1}$ that makes the diagram commute, and do something with the same flavor to get part (b).

Given a map $h: A \to C$, let's call the induced chain map a lift of $h$.

Apply this with $A=C$, also assuming that both chain complexes are exact and all of the $P_{i}$ and $Q_{i}$ are free: that is, assume that we have two free resolutions of $A$. Then we get lifts of the identity map $f: P_{*} \to Q_{*}$ and $g: Q_{*} \to P_{*}$. The composite $g \circ f$, as a map $P_{*} \to P_{*}$, is also a lift of the identity map of $A$ and so must be chain homotopic to the identity (since both it and the identity are lifts of the same map, they are chain homotopic by (b)), and similarly for $f \circ g$. Tensor with another abelian group $B$ and this will continue to hold: $(g \otimes 1_{B}) \circ (f \otimes 1_{B})$ will be chain homotopic to the identity map on the chain complex with terms $P_{i} \otimes B$, and the same for the other composite. Therefore both composites induces the identity on homology, which means that the maps $f \otimes 1_{B}$ and $g \otimes 1_{B}$ induce isomorphisms on homology.