# Characterization of quotient maps

The following is quoted from https://en.wikipedia.org/wiki/Quotient_space_(topology)

Quotient maps q : X → Y are characterized among surjective maps by the following property: if Z is any topological space and f : Y → Z is any function, then f is continuous if and only if f ∘ q is continuous.

Assuming that $$q$$ is a quotient map, I can prove the characterizing property using the definition of a quotient map, namely, $$q^{-1}U$$ is open if and only if $$U$$ is open. However, I could not see how to prove the converse: how does the property imply that $$q$$ is quotient?

Suppose $$q: X \to Y$$ obeys the "composition property". Suppose $$U$$ is a subset of $$Y$$ such that $$q^{-1}[U]$$ is open. Define $$g: Y \to \{0,1\}$$, where $$\{0,1\}$$ has the Sierpinski topology $$\{\emptyset,\{0\},\{0,1\}\}$$, by $$g(x) = 0$$ when $$x \in U$$ and $$g(x)=1$$ for $$x \notin U$$.

Then $$g \circ q: X \to \{0,1\}$$ sends $$q^{-1}[U]$$ to $$0$$ and all other points of $$X$$ to $$1$$ and because we assumed $$q^{-1}[U]$$ is open and as $$\{0,1\}$$ has the specified topology, $$g \circ q$$ is continuous. So by the "composition property", $$g$$ must be continuous, so $$g^{-1}[\{0\}]=U$$ must be open.

So we have show that the topology of $$Y$$ contains all $$U$$ such that $$q^{-1}[U]$$ is open and from this it follows that $$q$$ is a quotient map.

• I just realized through your argument that $Z$ can be chosen arbitrarily. Seems I misunderstood the logic. – Uchiha Jan 16 '19 at 16:54
• @Uchiha the power lies in that it holds for all spaces, although we use it only for Sierpiński space. – Henno Brandsma Jan 16 '19 at 16:57

Let $$\tau$$ denote the quotient topology on $$Y$$ induced by $$q$$.

Let $$\rho$$ denote a topology on $$Y$$ such that the characteristic property is satisfied.

Then we have actually two identity functions $$Y\to Y$$:

$$\mathsf{id}:(Y,\tau)\to(Y,\rho)$$ and $$\mathsf{id}:(Y,\rho)\to(Y,\tau)$$

Both can be shown to be continuous.

This implies that $$\tau=\rho$$.

edit (for further explaining)

In your question you mentioned that it was clear to you that the quotient topology induced by $$q$$ satisfies the characteristic property.

Then in order to show that this property is indeed "characteristic" it is enough to prove that the quotient topology is unique in this.

To do so I put forward a topology $$\rho$$ on $$Y$$ such that the property holds.

How it is defined does not really matter, and it is enough now to prove that this topology $$\rho$$ must be the same as quotient topology $$\tau$$.

The proof written out goes like this:

Function $$q:X\to Y$$ is continuous if $$Y$$ is equipped with the quotient topology $$\tau$$. For clarity let us denote here $$q_{1}:\left(X,\tau_{X}\right)\to\left(Y,\tau\right)$$ where $$\tau_{X}$$ denotes the topology on $$X$$.

Function $$q:X\to Y$$ is continuous if $$Y$$ is equipped with the quotient topology $$\rho$$. For clarity let us denote here $$q_{2}:\left(X,\tau_{X}\right)\to\left(Y,\rho\right)$$ where $$\tau_{X}$$ denotes the topology on $$X$$.

Be aware that $$q_{1}$$ and $$q_{2}$$ are both the function $$q$$. The thing that is different is the topology on its codomain $$Y$$.

Both setups carry the characteristic property.

Likewise we have functions $$\mathsf{id}_{1}:\left(Y,\tau\right)\to\left(Y,\rho\right)$$ and $$\mathsf{id}_{2}:\left(Y,\rho\right)\to\left(Y,\tau\right)$$ both prescribed by $$y\mapsto y$$ on set $$Y$$.

Then $$\mathsf{id}_{1}\circ q_{1}=q_{2}$$ so $$\mathsf{id}_{1}\circ q_{1}$$ is continuous.

From this we conclude that $$\mathsf{id}_{1}$$ is continuous on base of the characteristic property.

Also $$\mathsf{id}_{2}\circ q_{2}=q_{1}$$ so $$\mathsf{id}_{2}\circ q_{2}$$ is continuous.

From this we conclude that $$\mathsf{id}_{2}$$ is continuous on base of the characteristic property.

This implies that $$\tau=\rho$$.

• Sorry I could not follow your argument, in particular, how is $\rho$ precisely defined? It does not seem to make a particular choice of $Z$ as the other answer of Henno Brandsma? – Uchiha Jan 16 '19 at 16:55
• @Uchiha he takes $Z$ as $Y$ in the quotient topology, and $g$ the identity function between $Y$ and $Z=Y$. – Henno Brandsma Jan 16 '19 at 18:56
• In your question you mentioned that it was clear to you that the quotient topology induced by $q$ satisfies the characteristic property. Then in order to show that this property is indeed "characteristic" it is enough to prove that the quotient topology is unique in this. This is done in my answer. I put forward a topology on $Y$ that carries the property (how it is defined does not really matter). Then it is proved that this topology indeed coincides with the quotient topology. Tomorrow I will have a second look. Good night. – drhab Jan 16 '19 at 21:20
• I added an explanation. – drhab Jan 17 '19 at 8:59
• @drhab Thanks a lot for your detailed comments. I have the following questions:(a) How does one know that such a topology $\rho$ satisfying the characteristic property exists? (2) My undersdanding based Henno Brandsma's answer is that the arbitrariness of $Z$ plays an important role, but it does not seem to be reflected by your answer? (3) Basically $id=id_1=id_2$ (I understand you wanted to distinguish the topologies but mappings can be defined without topology). But isn't $id$ continuous w.r.t. any topology? So why can this be used for identifying topologies? – Uchiha Jan 17 '19 at 19:11