This is a homework problem I've been bashing my head against:

Consider a disjoint union of topological spaces $Z_i$ with the disjoint union topology, $$ Z = \bigsqcup_{i \in I} Z_i $$ Let $f:\ Z \twoheadrightarrow A$ be a surjection, and put the quotient topology generated by $f$ on $A$.

Prove that $U \subseteq A$ is open in $A$ $\iff$ for every $i \in I$, $U \cap f \left( Z_i \right)$ is open in $A$.

Here is what I've proven so far: (This may or may not be useful.)

  • $U \subseteq Z$ is open (closed) in $Z$ $\iff$ for every $i \in I$, $U \cap Z_i$ is open (closed) in $Z$.

  • The $( \Leftarrow )$ implication is fairly easy; if each $U \cap f \left( Z_i \right)$ is open, then we can write $U$ as a union of open sets: $U = \bigcup_{i \in I} U \cap f \left( Z_i \right)$. This implies that $U$ is open since topologies are closed under arbitrary unions.

  • The function $f$ is continuous just by definition of the quotient topology. ($U \subseteq A$ is open in $A \iff$ $f^{-1} \left( U \right)$ is open in $Z$.)

Now here's a problem: Consider $Z_1 = [0,\ 1]$, $Z_2 = [1,\ 2]$ with subspace topology inherited from the standard topology on $\mathbb{R}$. Let $f$ map $Z_1$ to $[0,\ 1]$ and $Z_2$ to $[1,\ 2]$. Identify the point $1$ in $Z_1$ and $Z_2$ as the same point in the quotient space $A = [0,\ 2]$. The function $f$ is essentially an identity map that glues the two intervals together, and is obviously surjective.

Now take $U = (1- \epsilon,\ 1+ \epsilon)$ for some small $\epsilon >0$. $U$ is open in $A$, since it's the image of the saturated open set $(1- \epsilon,\ 1] \cup [1,\ 1 + \epsilon)$. (This is open in $Z$ since it's a union of an open set in $Z_1$ and an open set in $Z_2$.)

Now $f \left( Z_1 \right) = [0,\ 1]$. This is not open in $A$ to the best of my knowledge, since $f^{-1} \left( f \left( Z_1 \right) \right) = Z_1 \cup \{ 1^\prime \}$, where $1^\prime \in Z_2$ is understood as different from the element $1 \in Z_1$, is not open in $Z$. Then $U \cap f \left( Z_1 \right) = (1-\epsilon,\ 1]$. This is not open in $A$.

What is the error in my reasoning here?


1 Answer 1


There is no error in your reasoning: the statement in the exercise is false.

No need to take $\epsilon$ very small: you can simply let $U=(0,2)$. There is a small error, presumably a typo, in your last paragraph: $f^{-1}\big[f[Z_1]\big]=Z_1\cup\{1'\}$, not $A\cup\{1'\}$. Indeed, you can show that $A$ that has its usual Euclidean subspace topology, so $U\cap f[Z_1]=(0,1]$ is certainly not open in $A$.

  • $\begingroup$ Thanks for that typo catch! $\endgroup$
    – Roger Burt
    Sep 17, 2015 at 6:51
  • $\begingroup$ @Roger: You’re welcome! $\endgroup$ Sep 17, 2015 at 6:51

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