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.
$$f_t \circ g = g \circ f_t \quad \forall t \in [0,1]?$$
Commuting Homotopies – Can One Find Commuting Homotopies of Maps?
algebraic-topologygeneral-topologyhomotopy-theory
Related Solutions
Let $p:[0,1] \to X$ be a loop based at $x_0$.
The map $h(s,t) = f_t(p(s))$ is a continuous mapping $[0,1] \times [0,1] \to X$.
On the boundary of the square, the following is true:
$h(0,t) = f_t(x_0) = h(1,t) \quad$ and $\quad h(s,0) = p(s) = h(s,1)$.
In other words, $h$ restricts to $p(s)$ along the bottom and top sides and restricts to $f_t$ along the left and right sides. The concatenation of two paths, $f_t*p(s)$, is the left side followed by the top side. The concatenation of two paths, $p(s)*f_t$, is the bottom side followed by the right side. That $h$ extends this pair of concatenated paths over the square shows that they are homotopic relative to their endpoints.
If you sketch this square and then draw a diagonal from the upper left to the lower left, then you can visualize the homotopy (rel endpoints) by restricting $h$ to the family of line segments which join the lower left vertex to a point on the diagonal and then to the upper right vertex.
Given a homotopy $H : A \times I \to B$ and $t \in I = [0,1]$ define $H_t : A \to B, H_t(a) = H(a,t)$.
Now let us denote your homotopy by $G * F$. This is certainly the most obvious solution because $(G * F)_t = G_t \circ F_t$, i.e. the homotopy $G * F$ is obtained by composing $G$ and $F$ levelwise.
Now let $\alpha, \beta : I \to I$ be continuous maps such that $\alpha(0) = \beta(0) = 0$ and $\alpha(1) = \beta(1) = 1$. Then $F^\alpha = F \circ (id_X \times \alpha)$ and $G^\beta = G \circ (id_Y \times \beta)$ are also homotopies between $h,h'$ and $k,k'$, respectively. Therefore also $G^\beta * F^\alpha$ is a homotopy as desired. But it definitely has no benefit to invoke $\alpha, \beta$ as additional ingredients.
Best Answer
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!