Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a set and let $(Y,\tau)$ be a topological space. For a given map $g:X\rightarrow Y$, define $$\tau'=\{U\subseteq X:U=g^{-1}(V) \text{ for some } V\in\tau\}.$$

Could anyone explain which of the following statements are/is true and which are/is false?

  1. $\tau'$ defines a topology on $X$.

  2. $\tau'$ defines a topology on $X$ only if $g$ is onto.

  3. Let $g$ be onto. Define an equivalence relation $x\sim y$ iff $g(x)=g(y)$. Then the quotient space of $X$ w.r.t. this $\sim$ with the topology inherited from $\tau'$ is homeomorphic to $(Y,\tau)$.

I must say, from the little I know about quotient topology, that $2$ is true. Thank you.

share|cite|improve this question
This is a special case of the final topology. If you know that $g^{-1}[\bigcup U_i]=\bigcup g^{-1}[U_i]$ and $g^{-1}[A\cap B]=g^{-1}[A]\cap g^{-1}[B]$, the proof of the first part should be easy. – Martin Sleziak Jan 20 '13 at 8:09
up vote 1 down vote accepted

Let’s see what we can say about $\tau'$.

  • $X=g^{-1}[Y]$ and $\varnothing=g^{-1}[\varnothing]$, so $X,\varnothing\in\tau'$.

  • Suppose that $U_0,U_1\in\tau'$. Then there are $V_0,V_1\in\tau$ such that $U_0=g^{-1}[V_0]$ and $U_a=g^{-1}[V_1]$, and $U_0\cap U_1=g^{-1}[V_0]\cap g^{-1}[V_1]=g^{-1}[V_0\cap V_1]\in\tau'$, since $V_0\cap V_1\in\tau$. Thus, $\tau'$ is closed under finite intersections. (Remember that inverse functions commute with unions and intersections.)

  • Suppose that $\mathscr{U}\subseteq\tau'$. Then for each $U\in\mathscr{U}$ there is a $V_U\in\tau$ such that $U=g^{-1}[V_U]$, and $\bigcup\mathscr{U}=\bigcup_{U\in\mathscr{U}}g^{-1}[V_U]=g^{-1}\left[\bigcup_{U\in\mathscr{U}}V_U\right]\in\tau'$, since $\bigcup_{U\in\mathscr{U}}V_U\in\tau$. Thus, $\tau'$ is closed under arbitrary unions.

We’ve just shown that $\tau'$ is a topology on $X$, irrespective of whether $g$ maps $X$ onto $Y$.

For the last question, let $Z=X/\sim$, and let $q:X\to Z:x\mapsto[x]$ be the quotient map, where $[x]$ is the $\sim$-equivalence class of $x$. Define $h:Y\to Z$ as follows:

For each $y\in Y$, $g^{-1}\big[\{y\}\big]$ is a $\sim$-equivalence class, by the definition of $\sim$. Specifically, for each $x\in X$ such that $g(x)=y$, $g^{-1}\big[\{y\}\big]=[x]\in Z$. Let $h(y)=g^{-1}\big[\{y\}\big]\in Z$.

To complete the argument, just show that $h$ is a homeomorphism. I’ll leave it to you to show that $h$ is one-to-one and onto. To show that $h$ is continuous, suppose that $U\subseteq Z$ is open in $Z$; we need to show that $h^{-1}[U]$ is open in $Y$, i.e., that $h^{-1}[U]\in\tau$. By the definition of $\tau'$ this is true if and only $g^{-1}\big[h^{-1}[U]\big]\in\tau'$. By the definition of quotient space $q^{-1}[U]\in\tau'$. But for any $x\in X$ we have $x\in q^{-1}[U]$ iff $[x]\in U$ iff $g(x)\in h^{-1}[U]$, i.e., $q^{-1}[U]=g^{-1}\big[h^{-1}[U]\big]$, and therefore $g^{-1}\big[h^{-1}[U]\big]\in\tau'$, as desired.

It only remains to show that $h$ is an open map. I’ll leave this to you: it requires no ideas beyond those that I used to show that $h$ is continuous.

share|cite|improve this answer

1 is true. To show so, check the axioms for topological spaces.

Obviously $\varnothing$ and $X$ are in $\tau'$.

Given a collection $\{U_\alpha \in \tau'\}$, we know that there exists $V_\alpha \in \tau$ for every $\alpha$ such that $U_{\alpha} = g^{-1}(V_\alpha)$. We also know that $\bigcup_\alpha U_\alpha = g^{-1}(\bigcup_\alpha V_\alpha)$. Since $\tau$ is a topology, $\bigcup_\alpha V_\alpha \in \tau$, and so $\bigcup_\alpha U_\alpha \in \tau'$.

Finally, $\bigcap_\alpha U_\alpha = g^{-1}(\bigcap_\alpha V_\alpha)$. If the collection is finite, then $\bigcap_\alpha V_\alpha \in \tau$ because $\tau$ is a topology, and so $\bigcap_\alpha U_\alpha \in \tau'$.

3 is also true. Let $q:X \to X/\sim$ be the quotient map. $X/\sim$ inherits the topology from $\tau'$, so $W \subseteq X/\sim$ is open $\Leftrightarrow$ $q^{-1}(W) \in \tau'$ $\Leftrightarrow$ $q^{-1}(W) = g^{-1}(V)$ for some $V \in \tau$. This suggests that $q \circ g^{-1}$ may be the homeomorphism we're looking for. I'll do this in the following steps:

  1. Prove that $q \circ g^{-1}$ actually defines a function from $Y$ to $X/\sim$. Since $g$ is onto, we know the domain can be $Y$. The only problem that may arise is $g^{-1}$ takes a point to a set, so $(q \circ g^{-1})(y)$ may be not be a single point in $X/\sim$. We show that this is not the case. Suppose $g^{-1}(x) = U$ is a set. Then for all $u \in U$, $g(u) = x$. By definition of $\sim$, $q(U)$ contains a single point.
  2. Prove that $q \circ g^{-1}$ is injective. Suppose $(q \circ g^{-1})(y_1) = (q \circ g^{-1})(y_2)$. Then there exist $x_1 \in g^{-1}(y_1)$ and $x_2 \in g^{-1}(y_2)$ such that $g(x_1) = g(x_2)$. But since $g(x_1) = y_1$ and $g(x_2) = y_2$, we must have $y_1 = y_2$.
  3. Prove that $q \circ g^{-1}$ is surjective. Suppose $z \in X/\sim$. Then there exists $y \in Y$ such that $q^{-1}(z) = \{x\ |\ g(x) = y\} = g^{-1}(y)$. It follows that $(q \circ g^{-1})(y) = z$.
  4. Show that $g \circ q^{-1}$ is continuous. Suppose $V \in \tau$. Then $g^{-1}(V) \in \tau'$. Let $W = (q \circ g^{-1})(V)$. Then $q^{-1}(W) = g^{-1}(V)$. By the definition of the induced topology of $X/\sim$, $W$ is open.
  5. Show that $q \circ g^{-1}$ is continuous. Suppose $W \subseteq X/\sim$ is open. There must exists $V \in \tau$ such that $g^{-1}(V) = q^{-1}(W)$, i.e., $V = (g \circ q^{-1})(W)$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.