Is bijection mapping connected sets to connected homeomorphism? If $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ is a bijection, mapping connected sets to connected, is $f$ necessarily a homeomorphism?
The converse is true, a well known property of homeomorphisms.
I know the result is true for $n = 1$. In that case, it's not very hard to see, that $f$ maps open intervals to open intervals: Image is going to be interval, and if it had a boundary point, some $x$ in the open interval would have to map to the boundary point. But now images of $(x, \infty)$ and $(-\infty, x) $ couldn't both be connected.
Also, pre-image of open interval is also open: If $x < y$ are pre-images of endpoints of the interval, the original interval is subset of image of $[x,y]$, but then clearly image of $(x,y)$ is the open interval.
Those two observations prove that $f$ is homeomorphism when $n = 1$. Connectedness is however a quite different phenomenon in higher dimensions, so I don't even have a clear idea if it should be true or not.
 A: Using the converse you mentioned, it is clear that if $f$ or $f^{-1}$ fails to map a connected set to another connected set, $f$ cannot be a homeomorphism. Thus, it suffices to show that 
Theorem: Fix $n > 1$. Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be a bijection, such that $f$ maps connected sets to connect sets, and $f^{-1}$ maps connected sets to connected sets. Then both $f$ and $f^{-1}$ are continuous. 
This is a corollary of Theorem 1 in a paper of Tanaka. I reproduce the proof below. 
Proof: since the hypothesis is symmetric in $f$ and $f^{-1}$, it suffices to prove that $f$ is continuous. We proceed by contradiction. Suppose $p\in \mathbb{R}^n$ is such that $f$ is discontinuous at $p$. Then there exists a sequence of points $(p_n)$ such that $p_n \to p$ and $f(p_n) \not\to f(p)$; without loss of generality we assume that $f(p) = 0$. Thus, up to a subsequence we can assume that there exists $\epsilon > 0$ such that $f(p_n) \in B_\epsilon^c = \mathbb{R}^n \setminus B_\epsilon$. Clearly $\{0\} \cup B_\epsilon^c$ has two connected components. On the other hand, $f^{-1}( \{0\} \cup B_\epsilon^c)$ is connected: by assumption $f^{-1}(B_\epsilon^c)$ is connected since $B_\epsilon^c$ is connected, and we have that every open neighborhood of $p = f^{-1}(0)$ intersects $f^{-1}(B_\epsilon^c)$. Thus we forced a contradiction.  
Remark: In the case $n  = 1$, $B_\epsilon^c$ is not connected. However, it has two connected components $B_\epsilon^{c+}$ and $B_\epsilon^{c-}$. The proof can be carried through if we replace every instance of $B_\epsilon^c$ with either $B_\epsilon^{c+}$ or $B_\epsilon^{c-}$: at least one of the two must contain infinitely many $f(p_n)$. 
A: I think that answer is no and introduce some example.
Assume that $h:\mathbb{R} \rightarrow \mathbb{R}$ is a function such that satisfies in intermediate value property (thus maps connected sets to connected) 
but is not continous. Now define $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ 
as $f(x,y) = (x,y+h(x))$. $f$ is an example that says NO to above question.
