$\require{AMScd}$ I've been reading about the Ext and Tor functors from a number of sources which all claim that the Tor functor provides a natural way of extending the exact sequence of abelian groups $A' \otimes D \to A \otimes D \to A'' \otimes D \to 0$ on the left (where $0 \to A' \to A \to A'' \to 0$ is exact). The sources do not explicitly state what they mean by natural: I take it to mean that a commutative diagram $$\begin{CD} 0 @>>> A' @>>> A @>>> A'' @>>> 0 \\ @. @V{f}VV @V{g}VV @V{h}VV \\ 0 @>>>B' @>>> B @>>> B'' @>>> 0 \end{CD}$$ (with exact rows) induces a commutative diagram $$\begin{CD} 0 @>>> \operatorname{Tor}(A',D) @>>> \operatorname{Tor}(A,D)@>>> \operatorname{Tor}(A'',D) @>>>A' \otimes D @>>> A \otimes D @>>> A'' \otimes D @>>> 0 \\ @. @V{\operatorname{Tor}(f,1)}VV @V{\operatorname{Tor}(g,1)}VV @V{\operatorname{Tor}(h,1)}VV @V{f \otimes 1}VV @V{g \otimes 1}VV @V{h \otimes 1}VV \\ 0 @>>> \operatorname{Tor}(B',D) @>>> \operatorname{Tor}(B,D)@>>> \operatorname{Tor}(B'',D) @>>>B' \otimes D @>>> B \otimes D @>>> B'' \otimes D @>>> 0. \end{CD}$$ To prove this, we need to choose free resolutions for each of the six groups in the first diagram and create a (3 dimensional) commutative diagram of exact sequences. Then we can apply $\otimes D$ to everything, and build the long exact sequence, and naturality follows from that.
Proving that the 3D diagram of free resolutions is commutative is giving me trouble. (I'm visualizing the diagram by imagining the free resolutions coming out of the computer screen in front of each of the six groups in the first diagram above.) It's clear to me that all the squares in the planes in front each of the maps $f$, $g$, and $h$ are commutative, as well as all the squares in the top and bottom planes. But what about the squares in the planes parallel to the screen? Do we have to choose the maps between free resolutions specifically to make this work, or is it automatic?