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.


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.

  • $\begingroup$ Thank you for reviewing ! $\endgroup$ – H_1317 Feb 7 at 21:35

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