Prove that quotient maps $q$ are characterized by $f:Y\to Z$ being continuous iff $f\circ q$ is

Question

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?

Answer 1

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.

Answer 2

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 $.

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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 $\tau=\rho$.