# Topology and quotient spaces involving inverse function

Let $X$ be a set and let ($Y, \tau)$ be a topological space. Let $g : X \to Y$ be a given map. Define $\tau'=\{U \subset X : U = g ^{−1} (V ) \text { for some } V \in \tau\}$. Which of the following statements are true?

a. $\tau^{\prime}$ defines a topology on $X$.

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

c. Let $g$ be onto. Define the equivalence relation $x \sim y$ if, and only if, $g(x) = g(y)$. Then the quotient space of $X$ with respect to this relation, with the topology inherited from τ', is homeomorphic to (Y, τ).

I think a. is true, but am unable to show it why especially because the map $g$ is not given to be continuous. Whether onto implies continuity is unable to be proved by me. As for third one, the claim seems true, but the exact line of reasoning is quite elusive to me. Any help. Thanks beforhand.

(a) has nothing to do with $g$ being continuous. Continuity doesn't even have any meaning here because $X$ doesn't have a topology defined on it.

To see it's a topology, just blindly follow the axioms using the definition of open set provided: if $U_i\in\tau'$ for $i$ in some index set $I$, then $U_i=g^{-1}(V_i)$ for $V_i\in\tau$. Then

$$\cup_i U_i=\cup_i g^{-1}(V_i)=g^{-1}(\cup_i V_i)$$

which is in $\tau'$ because $\cup_i V_i\in\tau$ by the definition of a topology. Use the same approach to show that $\tau'$ is closed under finite intersections, and that $\varnothing,X\in\tau'$.

Since we've now shown (a) is true, (b) is certainly false.

I'll tell you as much that (c) is true, but you should try to prove it yourself. A hint is that there is a very obvious map from $X/\mathord\sim$ to $Y$.

• Is the proof of $c$ the same as that for first law of isomorphism for groups? Commented Jan 18, 2017 at 9:37
• @vidyarthi the idea is certainly similar, but I don't know if I'd call it the "same". Commented Jan 18, 2017 at 13:11

(a) Is true. The topology $\tau'$ is just the minimal topology on $X$ that makes $g$ continuous. For this we need all sets of the form $g^{-1}[O]$ for $O \in \tau$ to be in the topology. The facts that $\tau$ is a topology and that $g^{-1}$ preserves unions and intersections, imply that these sets already form a topology.

This idea is used in the subspace topology as well:if $X$ would be a subset of $Y$ and $g(x) = x$ would be the standard embedding, sets of the form $g^{-1}[O] = O \cap X$ wouldform the subspace topology on $X$.

This plus the fact that surjectivity of $g$ is irrelevant in the proof of (a), shows (b) is false.

As to c) we introduce some notation, $X / \sim = \{[x]; x \in X\}$ is the quotient space (the set of classes) w.r.t. $\sim$ defined as $x \sim y$ iff $g(x) = g(y)$ (which one easily checks to be an equivalence relation by properties of $=$), and $q(x) = [x]$ is the standard quotient map that sends $x$ to its class, and $O \subseteq X /\sim$ is open iff $q^{-1}[O]$ is open in $X$ (so in $\tau'$).

To define a homeomorphism between $X/\sim$ and $Y$ when $g$ is surjective, the idea is obvious: define $\hat{g}([x]) = g(x)$, for $[x] \in X/\sim$.

This is well-defined because if we choose another representative for $[x]$, $[x] = [x']$ iff $x \sim x'$ iff $g(x) = g(x')$, the result of the definition is the same. This definition makes a diagram commutative: $\hat{g} \circ q = g$ by definition.

$\hat{g}$ is injective: suppose $\hat{g}([x]) = \hat{g}([x'])$ which means by the definition that $g(x) = g(x')$, and this says that $x \sim x'$ so $[x] = [x']$, as required.

$\hat{g}$ is surjective when $g$ is: if $y \in Y$ we pick $x \in X$ with $g(x) = y$, but then $\hat{g}([x]) = g(x) = y$ and $y$ is in the image of $\hat{g}$.

$\hat{g}$ is continuous: suppose $O \subseteq Y$ is open. Then $\hat{g}^{-1}[O]$ is open in $X/\sim$ iff (definition of quotient topology) $q^{-1}[\hat{g}^{-1}[O]]$ is open in $\tau'$, but $x \in q^{-1}[\hat{g}^{-1}[O]]$ iff $q(x) = [x] \in \hat{g}^{-1}[O]$ iff $\hat{g}([x]) = g(x) \in O$ iff $x \in g^{-1}[O]$, so $q^{-1}[\hat{g}^{-1}[O]] =g^{-1}[O]$ which is in $\tau'$ by definition.

$\hat{g}$ is open: suppose $O \subseteq X/\sim$ is open. This means by definition that $q^{-1}[O] \in \tau'$, so $q^{-1}[O] = g^{-1}[U]$ for some $U \in \tau$. Claim: $\hat{g}[O] = U$ (which is then open, as required). To see the claim: Suppose $y \in \hat{g}[O]$, then $y = \hat{g}([x]) = g(x)$ for some $[x] \in O$. But $[x] = q(x)$, so $x \in q^{-1}[O] = g^{-1}[U]$, but then $y =g(x) \in U$. And if $y \in U$, pick $x$ with $g(x) = y$. ($g$ is surjective), so $x \in g^{-1}[U]$, and so $x \in q^{-1}[O]$, which means that $[x] \in O$ and so $y = g(x) = \hat{g}([x]) \in \hat{g}[O]$, showing the second inclusion.

Quite straightforward, though a bit tedious checking..

• If this question asked in some screening test, (c) would kill the time. How do I able to check very quickly using any properties of Homeomorphism.$X/~$ is hard to imagine. please help me, sir.
– user464147
Commented Oct 16, 2017 at 9:04
• I doubt if (a) is true or not. Commented Jan 2, 2019 at 12:27
• I doubt if (a) is true or not. Consider (Y,tau)=(R,euclidean) and X =[0,1] . g be Identity map from [0,1] to R, certainly not onto. Now, X is in tau' if tau' is a topology on X. Since g is id map, g^(-1)([0,1]) =[0,1] which is not inside tau' since [0,1] is not inside tau. Hence X is not in tau implies tau is not a topology. Taking X=(0,1) instead of [0,1] would imply g doesnot need to be onto always to make tau' a topology though. Commented Jan 2, 2019 at 12:36
• @ChakSayantan $Y \in \tau$ so $g^{-1}[Y] = X$ is in $\tau'$. Both $[0,1]$ and $[0,1)$ are not in $\tau$ so their inverse images are not required to be in $\tau'$ by the definition. Commented Jan 2, 2019 at 17:43
• @HennoBrandsma it is not necessary that for all open U in \tau, g^{-1}(U) is in \tau'. What is said is if V is in \tau', there is some U in \tau so that g^{-1}(U)=V. Again I am not convinced why g^{-1}(Y) =X. g is not surjective at all. Commented Jan 3, 2019 at 4:08