Motivation for definition of quotient map and "passing to the quotient"

Question

I'm a bit confused by the definition of of a quotient map between topological spaces. What is the reason for constructing a topology in this way? I am familiar with quotients from algebra and even so with the idea of taking the quotient by an equivalence relation, but defining quotient maps in this way seems strange/artificial to me. Especially since I'm used to seeing quotients on the domain, but quotient maps induce the quotient topology on their image.

Also, could someone explain to me "what is going on" when we pass to the quotient? I understand the existence and uniqueness properties from a purely logical point of view but have no intuition on what passing to the quotient is actually achieving. It seems to me like we are constructing a unique continuous function given a quotient space and another continuous function (that satisfies some properties). Is that all? Again, what is the motivation? How are these constructions helpful?

Answer 1

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.

Answer 2

Let $N < G$ represent that the subgroup) $N$ is normal in $G$. Then we use one of the theorems on Homomorphisms ( i.e, we're dealing with Algebra here ) that given a homomorphism $h$ between groups $G,G'$ , and $N$ normal in $G;N'$normal in $G'$ , then when $h: G \rightarrow G'$ passes to the quotient (i.e., when it's constant on cosets), i.e., it's constant in equivalence classes, we get the induced map $h_*$

$h_* G/N \rightarrow G'/N'$

$[Ng] \rightarrow [h(Ng)]$

i.e., the class/coset of $g$ in $G/N$ is sent to the class of $h[gN ]$ ; the class of the image under $h$.

Specifically , in our case, we're dealing with quotients in(Co)Homology Vector Spaces. The groups involved (Cycles, Boundaries) are Abelian, so that every subgroup is normal in the group. Quotients are cosets consisting of (Co) Homology classes . Then a class in G/N (Cycles, Boundaries) is sent to the class of it's image under the map.