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Would anyone please have a look at my proof for the following statement?

Let $X$ and $Y$ be topological spaces and $p: X \rightarrow Y$ a surjective continuous map. Consider the following property:

(A): For every topological space $Z$ and every continuous function $f: X \rightarrow Z$ such that $f(x)=f(y)$ if $p(x)=p(y),$ there is a unique continuous $g$ such that $f=g \circ p$.

Show that $(\mathrm{A})$ is satisfied if and only if there is a homeomorphism $h: Y \rightarrow X / \sim$ to the quotient space $X / \sim$ with the equivalence relation $x \sim y \Leftrightarrow p(x)=p(y),$ such that $h \circ p$ is the projection $\operatorname{map} X \rightarrow X / \sim$ We say that $(\mathrm{A})$ is the universal property of the quotient topology.

Proof:

[$\Rightarrow$] Let $\Pi: X \to X / \sim$ be the projection map, i.e. $\Pi(x) = [x] := \{z \in X: p(x) = p(z)\}$. For every $x,y \in X$ such that $p(x) = p(y)$, by defintion of $\Pi$, we have $\Pi(x) = \Pi(y)$. Then by the universal property $(A)$, there exists an unique continuous $h: Y \to X / \sim$ such that $\Pi = h \circ p$.

Similarly, if $x, y \in X$ such that $\Pi(x) = \Pi(y)$, then by definition of $\Pi$, we have $p(x) = p(y)$. $\Pi$ is clearly a surjective map, and $p$ is continuous by assumption, so by the universal property, there exists an unique $g: X / \sim \ \to Y$ such that $p = g \circ \Pi$.

$g$ and $h$ are continuous inverses of each other, since we have: $$p = g \circ \Pi = g \circ (h \circ p) = (g \circ h) \circ p, \text{ i.e. } g \circ h = \mathbb{I}_Y, \text{ and}$$ $$\Pi = h \circ p = h \circ (g \circ \Pi) = (h \circ g) \circ \Pi, \text{ i.e. } h \circ g = \mathbb{I}_{X / \sim}.$$

[$\Leftarrow$] Let $Z$ be a topological space and $f: X \to Z$ a continuous function such that $f(x) = f(y)$ if $p(x) = p(y)$. Define $\varphi: X / \sim \ \to Z$ as: $\varphi([x]) = f(x)$. As $\varphi([x]) = \varphi \circ \Pi(x) = f(x)$, and as both $\Pi$ and $f$ are continuous, $\varphi$ is also continuous.

We check that $\varphi$ is well-defined. Assume for a contradiction that it is not. Then there exists $S \in X / \sim$ such that $S$ is mapped by $\varphi$ to both $z_1$ and $z_2$ in $Z$, where $z_1 \neq z_2$. Since $S \in X / \sim$, $S$ is of the form $\{x_1,x_2,... \in X: p(x_1) = p(x_2) = ...\}$. Since $S \mapsto z_1$, this implies $p(x_1) = p(x_2) = ... = z_1$, and, similarly, since $S \mapsto z_2$, this implies $p(x_1) = p(x_2) = ... = z_2$. But this means $p(x_1) \neq p(x_1)$, a contradiction.

Now define $g = \varphi \circ h$. As the composition of continuous functions, $g$ is continuous. $g$ is well-defined since both $\varphi$ and $h$ are well-defined, and for any $x \in X$: $$g \circ p(x) = \varphi \circ h \circ p(x) = \varphi \circ \Pi(x) = \varphi([x]) = f(x)$$

We show that $g$ is unique: assume $g': Y \to Z$ is a continuous function such that $f = g' \circ p$. For all $y \in Y$, we want to show that $g'(y) = g(y)$. Now since $p$ is surjective, there exists $x \in X$ such that $p(x) = y$. Then we have: $$g'(y) = g'(p(x)) = g' \circ p(x) = f(x) = g \circ p(x) = g(p(x)) = g(y)$$

ensbana
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1 Answers1

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Some comments:

  1. In the two parts of the proof of [$⇒$], it is not clear what universal properties you are referring to.
  2. The proof of the fact that $φ$ is well-defined seems to be unnecessarily twisted. It is enough to directly show that $[x] = [y]$ implies $f(x) = f(y)$.
  3. The claim that $g$ is continuous and the corresponding reasoning is missing in [$⇐$].
  4. The proof of uniqueness of $g$ is unnecessarily complicated given that we have already been right-cancelling surjections at the end of the proof of [$⇒$].
user87690
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  • Thanks for the comment. Regarding comment (1), the universal property I’m using is the one in the question, i.e. the property (A): I take $Z$ and $g$ in the definition of the property to be $X / \sim$ and $h$, respectively. – ensbana Dec 28 '20 at 14:10
  • Regarding comment (3), I’ve added two parts to show that $g$ is continuous: the first one is “Define $\varphi: X / \sim \ \to Z$ as: $\varphi([x]) = f(x)$. As $\varphi([x]) = \varphi \circ \Pi(x) = f(x)$, and as both $\Pi$ and $f$ are continuous, $\varphi$ is also continuous.”, and the second one is: “Now define $g = \varphi \circ h$. As the composition of continuous functions, $g$ is continuous.” – ensbana Dec 28 '20 at 14:12
  • Regarding comments (2) and (4), I realize that the arguments I’m using might a bit unnecessarily long. But I’ve just started out with category theory, and I want to use this opportunity to refine my understanding of some concepts. In general, would it be correct to show well-defineness of a function by showing that if one element has two images, then those two images must coincide? (sort of similar to showing the injectivity of a function: two elements and one image, then the two elements must coincide?). – ensbana Dec 28 '20 at 14:18
  • Regardless, thanks for the comments. I’ll write shorter arguments once I’m more comfortable with these concepts. – ensbana Dec 28 '20 at 14:18
  • Ad (1). Then the first part of $[⇒]$ is clear, but it is not completely clear why $Π:X \to X/{\sim}$ satisfies (A)… are you using the other implication here? Ad (3), the argument works becaure $Π$ is quotient (not only continuous), but this is essentially (A) for $Π$, which we are currently proving. So what fact about quotients do we know before proving this theorem? – user87690 Dec 28 '20 at 14:49
  • Yes. On that part, I’m showing that: (A) is satisfied $\Rightarrow$ there is a homeomorphism $h: Y \to X / \sim$ to the quotient space $X / \sim$ with the equivalence relation $x \sim y \Leftrightarrow p(x) = p(y)$, such that $h \circ p$ is the projection map $\Pi$. I set it up so that I can use property (A) on the spaces $X,Y$ and $X / \sim$. – ensbana Dec 28 '20 at 14:57
  • I can only think of the surjectivity of $\Pi$, and the property that $\Pi$ sends elements $x$ and $y$ with $p(x) = p(y)$ to the same equivalent class. I can’t see why the first is necessary in this argument, and I’ve already written the second in the first paragraph of my proof. Am I missing something here? – ensbana Dec 28 '20 at 17:50
  • You use the property (A) of $p$ and apply it to $Π$ to get $h: Y \to X/{\sim}$, which is correct, but to get $g: X/{\sim} \to Y$ you would need to use the property (A) of $Π$ and apply it to $p$. But how do we know that $Π$ has (A)? – user87690 Dec 29 '20 at 09:03
  • Since both $\Pi$ and $p$ are surjective, and we have $p(x) = p(y) \Leftrightarrow \Pi(x) = \Pi(y)$ by construction? – ensbana Dec 29 '20 at 09:57
  • Sure, we have a bijection between $Y$ and $X/{\sim}$ just considering the sets and functions alone (without topology). But why is $g: X/{\sim} \to Y$ continuous? – user87690 Dec 29 '20 at 10:32
  • I’m tempted to show that $\Pi$ is an open map, that way $g^{-1}(U)$ for an open set $U$ in $Y$ would be open in $X / \sim$, but I’m stuck. When I go about analyzing open sets in $X$ and $X / \sim$, I don’t know which topology we’re looking at. Actually this is a problem I frequently encounter, i.e. not knowing the sets are open or closed with respect to which topology. – ensbana Dec 30 '20 at 08:40
  • $Π$ is not necessarily an open map. It is only a quotient map. Since you are proving someting about quotients, you need to use the definition of the quotient somewhere (or something that you already know follows from the definition). The open sets of $X$ are given as the space $X$ is given. You should consider the quotient topology on $X/{\sim}$ and you should know what sets are open there. To get the big picture, you may look at the general construction of initial and final topologies and see that embeddings and quotients are particular examples. – user87690 Dec 30 '20 at 11:06