Combining homotopies

I have two homotopies $H,G:D^n \times I \to Z$ with $H(x,0) = f(x)$, $H(x,1) = f'(x)$, $G(x,0) = f'(x)$, $G(x,1) = g(x)$ for some maps $f,f',g:D^n \to Z$.

$G$ has the additionnal property of being constant (in $t$) on $S^{n-1}$, i.e $$G|_{S^{n-1} \times I} = f' |_{S^{n-1}} = g|_{S^{n-1}}.$$

Is it possible to combine them in a homotopy $H'$ from $f$ to $g$ in such a way that $H'$ actually coincides with $H$ on $S^{n-1}$ ? I.e such that $$H' |_{S^{n-1}\times I} = H |_{S^{n-1} \times I}.$$

I know how to combine them to get a homotopy from $f$ to $g$, and this homotopy will then do the exact same thing as $H$ on $S^{n-1}$, the only problem is that it will do it twice as fast. So I thought of "slowing down" the homotopy as $||x||$ approaches 1 (i.e as $x$ approaches $S^{n-1}$) but I do not know if that is a good idea...

Best, Olivier

The usual way of composing the homotopies $H$ and $G$ would be to define $$H_0(x,t)=\begin{cases}H(x,2t), & t\in[0,\frac{1}{2}]\\ G(x,2t-1), & t\in[\frac{1}{2},1]\end{cases}$$

In pictures, we're kind of gluing one square on top of the other like

but what you're saying we want is something like

[Edit: I just realised I mislabeled the $x$ axis in those images, they should be $D^n$, not $S^n$.]

and we can do this precisely because $G$ is constant on the boundary of the disk (the boundary of the disk is represented by the left hand side of the images). One such homotopy would be $$H'(x,t)= \begin{cases} H(x,\frac{2t}{1+\|x\|}), & t\in[0,\frac{1}{2}(1+\|x\|))\\ G(x,\frac{2}{1-\|x\|}(t-\frac{1}{2}(1+\|x\|)), & t\in[\frac{1}{2}(1+\|x\|),1),\: \|x\|\neq 1 \end{cases}$$

(I hope that's right. It was a lot harder to write down than I thought it would be). The idea at least is to just reparametrise $H_0$ along $I$ so that we're 'moving through' $H$ and $G$ at linear rates depending on $\|x\|$ in a continuously deforming way, and so that they meet on this 'diagonal boundary' according to $f'$. We need the constant condition for $G$ on $S^n$ to ensure that this is still a continuous function.

• Thanks, exactly what I was looking for! Dec 3, 2014 at 9:30
• It's not entirely clear to me what you're asking. You may want to just post the new question. Sep 11, 2020 at 12:05