Let $f:X \rightarrow Y$ continuous, open and exhaustive, if we suppose that $Y$ is connected and $f^{-1}(\{y\})$ is also connected for all $y \in Y$. Proof that X is connected.
I want to proof that and find examples showing that the open and exhaustive conditions are necessary but I don't know how to start, Ideas and suggestions are appreciated, many thanks!
On
A more general statement: If $f: X \to Y$ is a (surjective) quotient map and all fibres $f^{-1}[\{y\}]$ are connected (this property on is own is sometimes called a "monotone" function), and $Y$ is connected, then so is $X$.
Proof: If $t: X \to \{0,1\}$ is an arbitrary continuous function (where the 2 point space is discrete), then we can define $t': Y \to \{0,1\}$ by noting that if $y \in Y$, the set $f^{-1}[\{y\}]$ is connected, and so has a unique image point under $t$ (as the only non-empty connected subsets are singletons in a discrete space), and we define this to unique value be $t'(y)$. By construction, $t' \circ f = t$ and as $t$ is continuous and $f$ is quotient, we know that $t'$ is continuous. As $Y$ is connected, $t'$ is constant and thus $t$ is constant. It follows that $X$ is connected.
An open map is quotient (but we now also know it for closed maps, and more arbitary quotient maps too).
Lets assume $X$ is not connected. Then there exist $\emptyset \neq A,B\subseteq X$, open, disjoint such that $A\cup B = X$. Since $f$ is open and onto, $A':=f(A), B':=f(B)$ are open in $Y$ and their union is the whole space $Y$. Now we have that there is not $y\in A'\cap B'$, since it there was such a $y$, $f^{-1}(y)$ is connected, and thus $f^{-1}(y)$ is either a subset of $A$ or a subset of $B$. Thus $A', B'$ are disjoint, open sets and their union gives $Y$, but this is a contradiction since $Y$ is connected.