Trying to review topology for a prelim, I'm starting to wonder exactly what topological properties do quotient maps, usually given as $p: X \rightarrow Y$, preserve? I believe quotient maps preserve compactness and path connectedness, but I'm not sure how it can be proved. Also, is it true that quotient maps do not preserve simple connectedness and discreteness? If that is so, what would be some good counterexamples? I would appreciate any helpful input on this, thanks.
[Math] Passing to quotients via quotient maps preserving topological properties
general-topology
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The definition of a quotient map is not very enlightening, in my opinion. The intuition behind $X/\sim$ is "crushing the equivalence classes to points" inside of $X$. This is best seen through some examples:
The interval $[0,1]$ with the relation $0\sim 1$ gives the quotient $[0,1]/\{0,1\} \cong S^1$, the circle.
More generally, $D^n/\partial D^n \cong S^n$, where $D^n$ is the closed $n$-disk and $S^n$ is the $n$-sphere.
But how does this relate to the technical definition? First, $X/\sim$ as a set should not distinguish between two points in the same equivalence class. Thus, it's natural to take $X/\sim$ as the set of equivalence classes.
As for the topology, for the quotient $f: X\to X/\sim$ to be continuous, we must require that for any open set $U\subset X/\sim$, we have $f^{-1}(U)$ is open in $X$. But we want the topology of $X$ to entirely determine the topology of its quotient, so it's natural to define the open sets of $X/\sim$ to be precisely the subsets with open preimage in $X$. Thus, we recover the definition.
Edit: Now, what do we mean by "passing to the quotient"? This is relatively easy to understand if we think of the quotient space $X/\sim$ as crushing the equivalence classes to points: If we have a continuous map $f: X \to Y$ (for some arbitrary space $Y$) that is constant on some equivalence class $S$ (say $f$ maps points in $S$ to the point $y\in Y$), then we can think of $f$ mapping the entire equivalence class $S$ to $y$. Hence, we can see $f$ as a map on the quotient $f:X/\sim \to Y$ where $f$ maps the equivalence class $S$ to $y$. If $f$ is constant on each equivalence class, this gives a well-defined mapping, so $f$ "passes" or "descends" to the quotient.
Your question has two parts:
Are there further topological properties that are invariant under diffeomorphism?
The straightforward answer is this: Diffeomorphisms preserve exactly the same topological properties as homeomorphisms do; nothing more, nothing less.
The reason for this is essentially definitional: A topological property is, by definition, a property that is preserved by homeomorphisms. Since every diffeomorphism is a homeomorphism, every topological property is preserved by diffeomorphisms. And if a particular property of smooth manifolds is preserved by diffeomorphisms but not by homeomorphisms, then it's not a topological property.
The other half of your question was
Are there further non-topological properties that are invariant under diffeomorphism?
This is a more interesting question. But first you have to establish an appropriate category of spaces to work in -- diffeomorphisms are only defined between smooth manifolds, which are topological manifolds with an additional structure called a smooth structure. So the appropriate question to ask is whether there are non-topological properties of smooth manifolds that are preserved by diffeomorphisms. There are, but they're more subtle. For example, one such property for compact smooth manifolds is whether they bound (smoothly) parallelizable manifolds. This is one way that exotic spheres can be distinguished from each other.
Best Answer
By definition quotient maps are continuous, so they preserve any topological property that is preserved by continuous maps, including compactness and connectedness. They are rather badly behaved in most other ways, however. For example, consider the map
$$p:[0,2]\to\{0,1,2\}:x\mapsto\begin{cases} 0,&\text{if }x=0\\ 1,&\text{if }0<x<2\\ 2,&\text{if }x=2\;, \end{cases}$$
where $Y=\{0,1,2\}$ is given the quotient topology: $U\subseteq Y$ is open iff $p^{-1}[U]$ is open in $[0,2]$. Clearly the open sets in $Y$ are $\varnothing,\{0,1\},\{1\},\{1,2\}$, and $Y$, so $Y$ has the particular point topology with $1$ as the distinguished point. $[0,2]$ is a compact metric space, so it’s hereditarily normal. $Y$, on the other hand, is $T_0$ but has none of the higher separation properties.
In fact a quotient map preserves the $T_1$ property iff its fibres are closed. (The fibres of a map are the inverses of singletons.) You’ll find a little more information on preservation and non-preservation of topological properties here.