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 $y\in Y$?
Best Answer
Your map $g$ is a right inverse (or a section) for $f$ which means $f \circ g = id_Y$. In general $g$ is not continuous, even if all fibers $f^{-1}(y)$ are finite. Here is a counterexample.
Let $Y = \mathbb R$, $X = \mathbb R \times D$ with the discrete space $D = \{0,1\}$ and let $f : X \to Y, f(y,i) = y$, be the projection. This is a surjective, continuous open map. Define $$g : Y \to X, g(y) = \begin{cases} (y,0) & y \le 0 \\ (y,1) & y > 0 \end{cases}$$ Clearly $g$ is a right inverse for $f$. The set $U = \mathbb R \times \{0\}$ is open in $X$, but $g^{-1}(U) = (\infty,0]$ is not open in $Y$.