If we have a continuous function $g\colon X\to X$ with $g(Y)\subset Y$ and $h\colon Y\to Y$, where $h$ is the restriction of $g$ to $Y\subset X$, is there a continuous, surjective function $\pi\colon X\to Y$ such that $$ \pi\circ g = h\circ\pi? $$
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1Do You assume $g(Y)\subset Y$?, since You write $h:Y\rightarrow Y$ – Peter Melech May 07 '21 at 12:54
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Yes, I assume this. – Corona21 May 07 '21 at 13:00
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3One interesting special case to look at is when $g$ is the identity map. Even that's not trivial. – Arthur May 07 '21 at 13:02
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@Corona21 I suggest to include this assumption in the question. – Peter Melech May 07 '21 at 13:03
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Ok, I've done that. – Corona21 May 07 '21 at 13:04
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@Arthur its not even trivial if $g$ is constant. – C Squared May 07 '21 at 13:16
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@Arthur this is an interesting question, and i am leaning towards the side that there is a counter example – C Squared May 07 '21 at 13:24
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I also have the feeling that there is a counter-example... – Corona21 May 07 '21 at 13:24
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2Oh, are we just looking for a single counterexample? That shouldn't be difficult. I thought we wanted to classify when it worked and when it didn't. – Arthur May 07 '21 at 13:28
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Not always. Take $X=\mathbb{R}$ and $Y=\{0,1\}$. Take $g:X\to X$ as $g(x)=0$. Then $h:Y\to Y$ is given by $h(y)=0$. Note that $g$ is constant, so it is continuous and $g(Y)\subseteq Y$.
Any continuous function $\pi:X\to Y$ must be constant, so it cannot be surjective, and any surjective function $\pi:X\to Y$ cannot be continuous.
C Squared
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