Say $X$ is a topological space, and $f_0, f_1, g: X \to X$ are continuous self-maps. Let's assume that both $f_0$ and $f_1$ commute with $g$, i.e. $$f_i \circ g = g \circ f_i \quad \text{for} \quad i=1,2.$$ Assume furthermore that $f_0$ is homotopic to $f_1$. Is there any hope that there exists a homotopy $f_t:[0,1] \times X \to X$ which commutes with $g$, i.e. $$f_t \circ g = g \circ f_t \quad \forall t \in [0,1]?$$


1 Answer 1


Unfortunately, this does not work in general, here is a counterexample: Consider the maps $f_0,f_1,g: [-1,1] \to [-1,1]$ given by $f_0(x)=x$, $f_1(x)=-x$ and $g(x)=x^3$. Assume that $g \circ f_t = f_t \circ g$ for all $t \in [0,1]$. Then for any $t$ we have $$ f_t(1) = f_t(g(1)) = g(f_t(1)) = f_t(1)^3 $$ and thus we must have $f_t(1) \in \{0,1,-1\}$, i.e. $1$ is fixed by the homotopy $f_t$, since the set $\{0,1,-1\}$ is discrete. However, this contradicts the fact that $f_0(1) = 1 \neq -1 = f_1(1)$.

Under some strong assumptions this does work though. If $X$ is additionally a real vector space and $g$ an $\mathbb{R}$-linear map, we can use the straight-line homotopy $f_t := tf_0+(1-t)f_1$. We can check that $g$ and $f_t$ commute, namely \begin{align} f_t(g(x)) &= tf_0(g(x)) + (1-t)f_1(g(x))\\ &= tg(f_0(x)) + (1-t)g(f_1(x))\\ &=g(tf_0(x)+(1-t)f_1(x))\\ &= g(f_t(x)). \end{align} I hope this helps!

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    $\begingroup$ Nice example, thanks! $\endgroup$ Commented Dec 20, 2023 at 11:21

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