Definition of an induced homotopy in Hilton-Stammbach's book A Course in homological algebra
Studying Hilton-Stammbach's (2nd Ed), on page 170, it starts from a pair of chain maps, $\varphi, \varphi^{'}: C \rightarrow C' $, for which there is a homotopy, $\Gamma$. Next, a map $\Gamma_{\sharp}: C \otimes_ {\lambda} D$ ( with $\Lambda $ a P.I.D.) is defined from the latter, and it is shown to be a homotopy between $\varphi_{\sharp}, \varphi^{'}_{\sharp}: C\otimes_ {\Lambda} D \rightarrow C' \otimes_{\Lambda} D $ (Which are just $ \varphi \otimes 1, \varphi^{'} \otimes 1 $, respectively).
Here is my question: Right next, it says that it can be proved that the homotopy $\Gamma$ also induces a homotopy $\Gamma^{\sharp} $ between the induced morphisms $\varphi^{\sharp}, \varphi^{'\sharp}: Hom_{\Lambda}(C',D) \rightarrow Hom_{\Lambda}(C,D)$. However, how can I define such map $\Gamma^{\sharp}$? Or, in other words, how it would be "induced"?
I suppose that to prove that this induced homotopy is indeed that, I have to use the differential defined for the second complex of chains, analogously to how it is done above for the homotopy $\Gamma_\sharp$. Still I can't think of how I can define it.
Solution 1:
$\newcommand{\Hom}{\operatorname{Hom}}$ Suppose as you say that $\varphi,\varphi' : C\to C'$ are chain homotopic through $\Gamma: C\to C'$, meaning that $\varphi - \varphi' = d_{C'}\Gamma + \Gamma d_C$. First, let me just go over the details of the following claim.
Claim 1. For any other complex $D$, the resulting morphisms $\varphi\otimes 1,\varphi'\otimes 1 : C\otimes D\to C'\otimes D$ are homotopic via $H = \Gamma\otimes 1$.
Proof. There is a cancellation occurring because of the sign rule:
- The differential of $C\otimes D$ is $\delta = d_C\otimes 1 + 1\otimes d_D$, so we compute
$$(\Gamma \otimes 1)\delta = \Gamma d_C\otimes 1 + \Gamma \otimes d_D$$
- The differential of $C'\otimes D$ is $\delta' = d_{C'}\otimes 1 + 1\otimes d_D$, so we compute
$$\delta(\Gamma \otimes 1) = d_C\Gamma \otimes 1 + (1\otimes d_D)(h\otimes 1) \stackrel{!}=d_{C'}\Gamma \otimes 1 - \Gamma \otimes d_D.$$
This thus gives $\delta H +H\delta = (d_{C'}\Gamma +\Gamma d_C)\otimes 1$, and you can conclude. $\square$
I carried out this computation because something similar will happen when computing with $\Hom$, and in fact here the only degree $+1$ map $$\Hom(C',D) \longrightarrow \Hom(C,D)$$ you can possibly produce with your information is precomposition with $\Gamma$, usually denoted $\Hom(\Gamma ,1)$ or $\Gamma ^*$ (this is what you're writing with the sharp symbol).
Note! We have that $\Gamma ^*(f) = (-1)^{|f|} f\circ \Gamma $ since $\Gamma$ has degree $-1$ (sign rule!).
Claim 2. Precomposition with $\Gamma$ induces a homotopy $$\Gamma^* : \Hom(C',D) \longrightarrow \Hom(C,D)$$ between precomposition with $\varphi$ and precomposition with $\varphi'$.
Proof. Again, one has to be careful about cancellations and sign rules.
The differential of $\Hom(C,D)$ will be $\delta(f) = d_Df -(-1)^{|f|} fd_C$. These type of differentials have the property that they behave well for composition: for two composable maps of complexes $g : X\to Y$ and $f: Y\to Z$, and writing $\delta$ for the three possible differentials, we have that
$$\delta(f\circ g) = \delta(f)\circ g + (-1)^{|f|} f\circ \delta g.$$
With this at hand, we compute for a map $f : C' \to D$ that:
$$ \delta(\Gamma^*f) = (-1)^{|f|}\delta(f\circ \Gamma) = (-1)^{|f|}\delta(f)\circ \Gamma + (-1)^{|f|+|f|}f\circ\delta(\Gamma).$$
This has two summands:
$$\tag{1} (-1)^{|f|}(d_Df -(-1)^{|f|} fd_{C'})\circ \Gamma = (-1)^{|f|}d_Df\Gamma - fd_{C'}\Gamma$$ $$\tag{2} f\circ\delta(\Gamma) = f\circ (d_{C'}\Gamma + \Gamma d_C).$$
Note this last summand is $f\circ (\varphi-\varphi') = (\varphi-\varphi')^*(f) = \varphi^*(f) -\varphi'^*(f)$.
At the same time, since $\delta(f)$ has degree $|f|-1$ we compute that
$$\tag{3} \Gamma^*\delta(f) = (-1)^{|f|-1}\delta(f)\circ \Gamma = (-1)^{|f|-1}(d_Df-(-1)^{|f|} f\circ d_{C'})\circ \Gamma$$ $$= -(-1)^{|f|}d_D f \Gamma + f d_{C'} \Gamma.$$
Examining thing closely, you can see that $(1)$ and $(3)$ cancel, and all we are left with is the equation
$$\Gamma^*(\delta f)+ \delta \Gamma^*(f) = \varphi^*(f) -($\varphi')^*(f),$$ which says precisely that $\Gamma^*$ is a homotopy between $ \varphi^*$ and $ (\varphi')^*$. $\square$
Phew! Now you can try and do it for the other coordinate in $\Hom$.