The converse of continuous image of a connected set is connected

Question

I have recently proven that given $f:X\rightarrow Y, f$ continuous, X connected, then Y is connected. I wonder if the converse is true if we consider open mapping. In other words, Given X and Y connected set, if $\exists f:X\rightarrow Y$ s.t. $f(U)$ is open in Y $\forall$ open set $U\subset X$, then can we prove that f is continuous?

Answer 1

No: let $f$ be the identity map from $\mathbb{R}$ with the trivial topology $\{\emptyset,\mathbb{R}\}$ to $\mathbb{R}$ with its usual topology.

Answer 2

Not in general. Take $X=S^1$ and $Y=[0,2\pi)$, both endowed with their usual metrics. If you see $S^1$ as the unit circle in $\mathbb R^2$, then each element of $S^1$ can be written in one and only one way as $(\cos\theta,\sin\theta)$, with $\theta\in[0,2\pi)=Y$. Now, define $f\colon X\longrightarrow Y$ be $f(\cos\theta,\sin\theta)=\theta$. It is an open map, but it is not continuous: if $c_n=\left(\cos\left(2\pi-\frac1n\right),\sin\left(2\pi-\frac1n\right)\right)$, then $\lim c_n=(1,0)$, but the sequence $\bigl(f(c_n)\bigr)_{n\in\mathbb N}$ doesn't converge in $Y$.