# Quotient map and connected sets proof

I'm hoping to have my proof reviewed for correctness. Thanks in advance!

Let $$p: X \to Y$$ be a quotient map. Show that if each set $$p^{-1}(\{y \})$$ is connected, and if $$Y$$ is connected, then $$X$$ is connected.

Proof:

Let $$p$$ be a quotient map from $$X$$ to $$Y$$.

Since $$p$$ is a quotient map we know that $$p$$ is surjective.

Suppose $$X$$ is not connected.

Then each $$X_i = p^{-1}(\{y_i\})$$ is a connected set in a separable space X. Let the open sets $$A$$ and $$B$$ satisfy the condition that $$A \cup B = X$$, where $$A$$ and $$B$$ are disjoint and nonempty.

Then each $$X_i$$ must be fully contained within either $$A$$ or $$B$$, since each $$X_i$$ is connected.

Therefore for all $$x_i{_a} \in A$$ we first have that $$\bigcup x_i{_a} = A$$, since we've taken each $$x_i$$ stemming from every $$y \in Y$$.

Second, $$p(\bigcup x_i{_a})$$ maps to an open set in Y since $$\bigcup X_i{_a}$$ is open and the complete reverse image of a quotient map. Similarly, $$p(\bigcup X_i{_b})$$ maps to an open set in $$Y$$ also.

These sets in $$Y$$ must be disjoint because if they were not disjoint that would imply that some $$x \in X_i$$ belonged to both $$A$$ and $$B$$. But then $$p(x)$$ would map to a point $$y$$ that has a pre-image in two disjoint sets -- since the pre-image $$p^{-1}(\{y \})$$ is only one connected set, so it cannot occupy space in both $$A$$ and $$B$$. So we've shown the image sets are disjoint.

And $$p(\bigcup X_i{_a}) \cup p(\bigcup X_i{_b}) = Y$$ since every quotient map is surjective and $$A \cup B = X$$.

Hence we have shown $$Y$$ to have a separation. A contradiction. Hence $$X$$ must be connected.

Your proof is correct. However, the phrase "in a separable space $$X$$" is misleading. A separable space is one which has a countable dense set. You mean a space which has a separation into two nonempty disjoint open subsets.