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.

  • $\begingroup$ 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. $\endgroup$ Commented Jan 20, 2013 at 8:09

2 Answers 2


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.


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., $V = (g \circ q^{-1})(W)$.

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