There's a proof of the uniqueness of the quotient topology here : Uniqueness of the Quotient Topology

Now I almost want to accept this proof however, there are two issues. First let me state some theorems and definitions

Characteristic Property of the Quotient Topology : Suppose $X$ and $Y$ are topological spaces and $q : X \to Y$ is a quotient map. For any topological space $Z$, a map $f : Y \to Z$ is continuous if and only if $f \circ q $ is continuous

Uniqueness of the Quotient Topology : Given any topological space $X$, a set $Y$ and a surjective map $q : X \to Y$, the quotient topology is the only topology on $Y$ for which the characteristic property

Now the book which I am using, Introduction to Topological Manifolds, defines a quotient map in the following way.

Definition of Quotient Topology: Let $X$ be a topological space, $Y$ be any set, and $q : X \to Y$ be a surjective map. Define a topology on $Y$ be declaring a subset $U \subseteq Y$ to be open if and only if $q^{-1}[U]$ is open in $X$. This is called the quotient topology induced by the map $q$.

Definition of a Quotient Map: If $X$ and $Y$ are topological spaces, a map $q : X \to Y$ is called a quotient map if it is surjective and $Y$ has the quotient topology induced by $q$.

Now onto my questions,

Question 1: In proving the uniqueness of the quotient topology, as done in the proof I linked, one would assume there exists another topology, say $\mathcal{T}$, on $Y$ for which the characteristic property holds.

But if you look at the characteristic property one of the requirements is that some map $q : X \to Y$ is a quotient map, but by the definition of a quotient map, $\mathcal{T}$ then has to be be the quotient topology (if it wasn't we couldn't talk about $q$ being a quotient map).

Is this observation correct?

Question 2: I'm assuming that the above isn't the case (otherwise the above two paragraphs would be the whole proof), so if I'm wrong (and if the post I linked to isn't using different definitions), then in the proof I linked to, why hasn't the map $q'$ been checked to see if it is actually a quotient map (otherwise the characteristic property wouldn't hold).

For example, in the accepted answer the following is said : "To show that $f_2$ is continuous first use the forward direction of the property of $\tau_d$ to show that $q'=f_3\circ q'$ is continuous (as you have done). Then noting that $q'=f_2\circ q$ use the characteristic property of the quotient space $Y_q$ to conclude that $f_2$ is continuous." But if we don't know if $q'$ is a quotient map or not, then we can't use the characteristic property for $\tau_d$.

Furthermore if $q'$ is a quotient map, then my above argument in the above paragraph leads me to believe that $\mathcal{T_d}$ has to be the quotient topology, which would make the rest of the proof in the link redundant.

  • $\begingroup$ Apologies for the length of this post, but it is the only way to put across all my arguments. $\endgroup$ – Perturbative Oct 1 '17 at 2:26
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    $\begingroup$ The proper statement that you should prove to clear up all this confusion is this: A surjective, continuous map $q:X \to Y$ is a quotient map $\iff$ (for all maps $f:Y \to Z$, ($f \circ q$ is continuous $\iff f$ is continuous)). $\endgroup$ – Alex Provost Oct 1 '17 at 4:47
  • $\begingroup$ @AlexProvost: You even don't need to assume continuity of $q$ before the equivalence. $\endgroup$ – user87690 Oct 1 '17 at 8:53
  • $\begingroup$ Your observation for question 1 is correct. However what you want to prove is that the quotient topology is the only topology satisfying the characteristic property given before $\endgroup$ – Julien Oct 1 '17 at 10:11
  • $\begingroup$ @user87690 You're right: taking $f$ to be the identity map of $Y$, which is continuous, yields that $q$ is continuous for free. $\endgroup$ – Alex Provost Oct 1 '17 at 13:43

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