Projecting Graph of a Function acted on by a homeomorphism

Question

Let $X,Y$ be compact, connected, simply-connected, and separable, metric spaces each with at-least $2$-points, and let $f,g:X\rightarrow Y$ be continuous functions. Does there always exist a homeomorphism $\Phi:X\times Y \rightarrow X\times Y$ such that $$ g(x) =\pi_Y\circ \Phi(x,f(x)) $$ for all $x \in X$, where $\pi_Y$ is the canonical projection of $X$ onto $Y$?

Heuristically (but not exactly): "Can we always perturb the graph of a continuous function so that it becomes the graph of another function?"

Dejan Govc · Accepted Answer · 2020-08-17T11:30:14.847

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No. Take $I=[0,1]$, $Q=[0,1]\times[0,1]$ and $J=\{0\}\times[1,2]$. Let $X=I$ and $Y=Q\cup J$. Let $f:X\to Y$ be the constant function $f(x)=(0,\frac32)$ and let $g$ be the constant function $g(x)=(\frac12,\frac12)$. Then any homeomorphism $\Phi$ of $X\times Y$ will preserve the square $I\times J$, so $\pi_Y(\Phi(x,f(x)))\in J$ and therefore cannot be equal to $g(x)$.

Added a bit later: The answer is negative even if you additionally assume that the two spaces are homogeneous. Take $X=Y=S^2$. Let $f:X\to Y$ be the identity map of $S^2$ and let $g:X\to Y$ be a constant map taking the value $c\in S^2$. Then the requirement $g(x)=\pi_Y(\Phi(x,f(x)))$ becomes $\pi_Y(\Phi(x,x))=c$. In other words, $\Phi(x,x)=(h(x),c)$ where $h:S^2\to S^2$ is a homeomorphism. But $(S^2\times S^2)\setminus\Delta$ (here $\Delta$ is the diagonal) and $(S^2\times S^2)\setminus (S^2\times\{c\})$ are not homeomorphic, so there is no such $\Phi$.

edited Aug 17 '20 at 11:30

answered Aug 17 '20 at 10:40

Dejan Govc

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What if $X=Y=\mathbb{R}^n$? Would things workout better? – ABIM Aug 18 '20 at 11:12
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Yes, then you can take $\Phi(x,y)=(x,y-f(x)+g(x))$. – Dejan Govc Aug 18 '20 at 15:59
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You could also e.g. replace $\mathbb R^n$ by a Lie group in which case $\Phi(x,y)=(x,yf(x)^{-1}g(x))$ would do the job. – Dejan Govc Aug 18 '20 at 16:08
Do we really need continuity of both $f$ and $g$ in this case.. I guess not? – ABIM Aug 19 '20 at 14:49
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@James_T: Well, mainly to ensure $\Phi$ is continuous. But if the discontinuities of $f$ and $g$ "cancel out", this still works, I guess. – Dejan Govc Aug 19 '20 at 16:41

Projecting Graph of a Function acted on by a homeomorphism

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