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Suppose $X,Y$ are topological spaces with $Y$ homogeneous and $f,g:X\to Y$ continuous such that there exist continuous functions $u,v:Y\to Y$ such that $$f = u\circ g \text{ and } g= v \circ f.$$

Does this imply that there is a homeomorphism $\varphi:Y\to Y$ such that $f = \varphi \circ g$?

Note. When the homogeneity condition on $Y$ is dropped, there are counterexamples: Let $y,z\in Y$ such that there is no homeomorphism on $Y$ sending $y$ to $z$, and let $f$ be the constant function from $X$ sending everything to $y$, and let $g$ be the constant function from $X$ sending everything to $z$.

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Counterexample. Take $X=\{0,1,2,3\}$ to be the discrete space with four points and $Y=S=\{e^{it}; 0\le t<2\pi\}$ to be a circle. Then you can simply take different orderings or the images. (Or, to be more precise, sufficiently different orderings. This will not work, for example, if you simply shift the images cyclically.)

For example:
$f(0)=1$, $f(1)=e^{i\pi/2}$, $f(2)=e^{i\pi}$, $f(3)=e^{i3/2\pi}$
$g(0)=f(0)$, $g(1)=f(1)$, $g(2)=f(3)$, $g(3)=f(2)$

Then you are asking about a homeomorphism $\varphi\colon S\to S$ such that \begin{align*} \varphi(1)&=1\\ \varphi(e^{i\pi/2})&=e^{i\pi/2}\\ \varphi(e^{i\pi})&=e^{i3\pi/2}\\ \varphi(e^{i3\pi/2})&=e^{i\pi} \end{align*}

From the values in $e^{i\pi/2}$ and $e^{i\pi}$ you see that images of the points between them (i.e., the image of the set $\{e^{it}; t\in[\pi/2,3\pi/2]\}$) cover the arc between $e^{i\pi/2}$ and $e^{i3\pi/2}$, which implies that one of the values $e^{i\pi}$ and $1$ is attained on this interval. Therefore $\varphi$ cannot be bijective.

On the other hand, it is not difficult to find a continuous function $S\to S$ with any prescribed values in these four points, therefore $u$ and $v$ with the required properties exist.

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