# Can one find commuting homotopies of maps?

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. $$f_t \circ g = g \circ f_t \quad \forall t \in [0,1]?$$

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!