Here is Prob. 11, Sec. 23, in the book Topology by James R. Munkres, 2nd edition:

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

My Attempt:

By definition, $p$ is a surjective map of the topological space $X$ onto the topological space $Y$, such that, for each set $V \subset Y$, the set $p^{-1}(V)$ is open in $X$ if and only if $V$ is open in $Y$.

Now suppose that $X$ is not connected. Then, by definition, we can write $X = C \cup D$, where the sets $C$ and $D$ are non-empty, disjoint, and open in $X$; that is, let the sets $C, D$ constitute a separation of $X$.

Let $y \in Y$. As the set $p^{-1}(\{ y \})$ is connected, so by Lemma 23.2 in Munkres, we must have either $p^{-1}(\{ y \}) \subset C$ or $p^{-1}(\{ y \}) \subset D$, but not both since $C \cap D = \emptyset$.

In particular, for every point $x \in X$, as $p(x) \in Y$, so we must have either $p^{-1}( \{ p(x) \} ) \subset C$ or $p^{-1}( \{ p(x) \} ) \subset D$, but not both.

Am I right?

Let $U$ and $V$ be the following subsets of $Y$. $$ U \colon= \{ \ y \in Y \ \colon \ p^{-1}(\{ y \}) \subset C \ \}, \qquad V \colon= \{ \ y \in Y \ \colon \ p^{-1}(\{ y \}) \subset D \ \}. $$ Then $U$ and $V$ are disjoint sets, because $C$ and $D$ are disjoint.

Am I right? Is the very last assertion in need of further elaboration?

Moreover, since $p$ is surjective and since $C \cup D = X$, we can conclude that $Y = U \cup V$.

Am I right? If so, then do I need to be any more explicit as to why this is so?

We now show that $p^{-1}(U) = C$ and $p^{-1}(V) = D$.

Let $x \in p^{-1}(U)$. Then $p(x) \in U$, and so $p^{-1}( \{ p(x) \} ) \subset C$, by the definition of set $U$. But as $x \in p^{-1}( \{ p(x) \} )$, so $x \in C$ also. Thus we have $p^{-1}(U) \subset C$.

Is my reasoning correct?

Conversely, suppose that $x \in C$. As in the fourth paragraph of this proof, we can conclude that either $p^{-1}( \{ p(x) \} ) \subset C$ or $p^{-1}( \{ p(x) \} ) \subset D$, but not both. But $x \in C \cap p^{-1}( \{ p(x) \} )$. So we must have $p^{-1}( \{ p(x) \} ) \subset C$, which implies that $p(x) \in U$ and hence that $x \in p^{-1}(U)$. Thus $C \subset p^{-1}(U)$.

Is this part of my reasoning correct and clear enough too?

From the preceding two paragraphs, we can conclude that $C = p^{-1}(U)$. Similarly, we can also show that $D = p^{-1}(V)$.

Now as $C = p^{-1}(U)$ and $D = p^{-1}(V)$ are open in $X$ and as $p$ is a quotient map, so the sets $U$ and $V$ are open in $Y$.

As $p$ is surjective, so we have $$ p(C) = p\left( p^{-1}(U) \right) = U, $$ and similarly also $p(D) = V$.

Now as $C$ and $D$ are assumed to be non-empty and as $p(C) = U$ and $p(D) = V$, so both $U$ and $V$ are non-empty as well.

Thus we have shown that $U$ and $V$ are non-empty disjoint open sets in the topological space $Y$ such that $Y = U \cup V$. Thus the sets $U, V$ constitute a separation of $Y$, which contradicts our hypothesis that $Y$ is connected.

Thus our supposition that $X$ is not connected is wrong. Hence $X$ is connected.

Is this proof correct? If not, then at which point(s) is (are) there problems therein?

  • $\begingroup$ To me, everything seems to be perfectly clear and correct. $\endgroup$ – Peter Elias Dec 12 '17 at 21:06
  • $\begingroup$ It all looks good, but wow, that's a lot to write! The key idea is to write $X=C\cup D$, as you did. Then the crux is to show $p(C)$ and $p(D)$ are open in $Y$. This follows by noting (as you did) that, if $y\in p(C)$, then $p^{-1}(y)\subset C$. That is, $C$ is saturated, and hence $p(C)$ is open. $Y=p(C)\cup p(D)$ and you're done. $\endgroup$ – Steve D Dec 12 '17 at 21:15

I really enjoyed reading this! This has a good level of detail.

The only part that I got stuck was where you said

Moreover, since $p$ is surjective and since $C \cup D = X$, we can conclude that $Y = U \cup V$.

This sentence alone is not enough to justify the conclusion.

Based on the definitions of $U,V$ you want to establish that for every $y\in Y$, either $p^{-1}(\{y\})\subseteq C$ or $p^{-1}(\{y\})\subseteq D$. But you already showed this in a previous part of your post!

So in other words, you should have said

Moreover, we already showed that if $y\in Y$ then $p^{-1}(\{y\})\subseteq C$ or $p^{-1}(\{y\})\subseteq D$. So, if $y\in Y$ then $y\in U\cup V$ and therefore $Y = U\cup V$.


An alternative way (more category theory in taste): suppose $f: X \to \mathbf{2}:= \{0,1\}$ is continuous (where the latter set has the discrete topology).

For any $y \in Y$, $p^{-1}[\{y\}]$ is connected and so $f[p^{-1}[\{y\}]]$ is either $\{0\}$ or $\{1\}$ (these are the only connected subsets of $\mathbf{2}$). We define $\tilde{f}(y)$ to be that value.

Note that $\tilde{f} \circ p = f$ by definition and so by the standard characterisation of quotient maps by composition properties, $\tilde{f}$ is continuous, as $f$ is, and so $\tilde{f}$ is a constant map as $Y$ is connected. But then $f$ is also a constant map and thus $X$ is connected.

Your argument is fine too, and is a more verbose version of what I wrote above, if you think about it. The idea is the same.


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