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Bit of a silly question, but here it goes:

We know that if a bijective continuous map is open, its inverse is continuous. But now suppose you have a surjective, open continuous map $f:X\rightarrow Y$. If I just pick an element $z\in f^{-1}(y)$ for every $y$ and then define $g:Y\rightarrow X$ with $g(y)=z$, will this $g$ be continuous (and be an inverse to $f$)? If not, is this possible if I add the constraint that $f^{-1}(y)$ is finite for every $y\in Y$?

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  • $\begingroup$ $g^{-1}(A)=f(A)$ by elementary set theory. Thus $g$ is continuous... $\endgroup$ May 17 '19 at 11:01
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Your map $g$ is a right inverse (or a section) for $f$ which means $f \circ g = id_Y$. However, if $f$ is not bijective, then $g \circ f \ne id_X$, so it is not a left inverse.

To see that $g$ is continuous, let $U \subset X$ be open. Then $g^{-1}(U) = g^{-1} (f^{-1}(f(U))) = (f \circ g)^{-1}(f(U)) = f(U)$ is open in $Y$. Note that in general $g$ is not an open map. In fact, in general not even $g(Y)$ will be open in $X$.

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