I Meaning of "passage to the quotient"?

[elias001]

TL;DR Summary

Would like to know the meaning of the phrase "passage to the quotient" and how is used in various mathematical contexts.

The following are taken from

Locally Compact Groups by Markus Stroppel

Commutative Algebra Volume 1 By Oscar Zariski, Pierre Samuel

Introduction to the Algebraic Geometry and Algebraic Groups Volume 39 by Michel Demazure

Basic Algebra by Anthony Knapp

General Topology by J. Dixmier

Commutative Algebra: Constructive Methods Finite Projective Modules by Henri Lombardi, Claude Quitte


Background

An inexperienced reader might conjecture that semigroup ideals essentially describe all semigroup homomorphisms, just as ring ideals (or normal subgroups) essentially describe ring (or group) homomorphisms as $\color{blue}{\text{passage to the quotient}}$ by its kernel.

b) Again let $A$ be arbitrary; let $(a_j)_{j\in J}$ be a family of elements of $A$ and let $a$ be the ideal of $A$ generated by these elements. Let $M$ be the $(A/a)-$module $\Omega_{A/k}/(a\Omega_{A/k}+\sum_j Ada_j)$ and $D:A/a\to M$ the map derived from $d$ by $\color{blue}{\text{passage to the quotient}}$ Then $(M,D)$ is a solution of the...

33. Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{K}$, and let $\iota$ be the linear map obtained as the composition of $\mathfrak{g}\to T^1(\mathfrak{g})$ and the $\color{blue}{\text{passage to the quotient}}$ $U(\mathfrak{g})$. Prove that $(U(\mathfrak{g}),\iota)$ has the following universal mapping property:....

...Methodologically, they are devoted to the decryption of different variations of the local-global principle in classical mathematics. For example, the localization at every prime ideal, the $\color{blue}{\text{passage to the quotient}}$ by every maximal ideal or the localization at every minimal prime ideal, each of which applies in particular situations.

We consider now the residue class ring $R/\mathfrak{a}$ of $R$ and we denote by $f,f'$, and $h$ the canonical homomorphisms of $R$ onto $R/\mathfrak{a}$, of $R_M$ onto $R_M/\mathfrak{a}^e$ and of $R$ into $R_M$ respectively. Since $\mathfrak{a}\subset \mathfrak{a}^{ec}$, $h$ defines, by $\color{blue}{\text{passage to the residue classes}}$, a homomorphism $\bar{h}$ of $R/\mathfrak{a}$into$R_M/\mathfrak{a}^e$which satisfies the relation$hf'=f\bar{h}##.

3.4.5 Example. Denote by $U$ the set of complex numbers of absolute value $1$. One knows that the mapping $x\mapsto g(x)=\mathrm{exp}(2\pi x)$ of $\mathbb{R}$ into $U$ is surjective, and that $g(x)=g(x')\leftrightarrow x-x'\in \mathbb{Z}$. Thus, if $p$ denotes the canonical mapping of $\mathbb{R}$ onto $T$, then $g$ defines, by $\color{blue}{\text{passage to the quotient}}$, a bijection $f$ of $T$ onto $U$ such that $f\circ p=g$.

Questions

The passages quoted above either contain the phrase "passage to the quotient" or "passage to the residue classes" (highlight in blue) in the context of quotient objects. I have taken from various sources at various levels. I would like to know how either of the phrase is used and what it means. More specifically, the phrase "the passage to" is confusing. I understand the process of quotienting out or modding out elements from a mathematical objects say a ring or a group. In the case of rings, modding out an element means sending that element to zero, for groups, the equivalent would be sending that element to the identity in the case of quotient group. But it does not make sense in the case of say point set topology for quotient topology. However common in all three of these different cases are partition of an object into blocks where elements from the same blocks are consider to be identical. Does than "the passage to a quotient" means to undertake such process. I know there is perhaps an attempt at conveying some sort of figure of speech imagery when using the phrase "the passage to". I could be wrong, and hence why I am asking.

Also in one of my tags, there doesn't seem to be a tag for math phrases, so i used Definitions"

Thank you in advance

 

[fresh_42]

Say we have a homomorphism, whether of a vector space, a group, a ring, an algebra, or a topological space, each of them with a different definition of what homomorphism means in the corresponding category, say $\pi \, : \, R\twoheadrightarrow R/I ,$ then passing to the quotient means applying this homomorphism. It means considering the equivalence classes in $R/I$ instead of the elements in $R.$ You could write those classes as $r+I$ (or possibly $rI$ in the case of groups), but it is cumbersome always having to write $r+I$ or writing $\pi(r)$ without gaining any new information. So these equivalence classes are often written
$$
\bar{r}=\pi(r)=r+I
$$
and even the bar is occasionally dropped. Authors could say passing to the quotient and keep writing the new elements which are now cosets as $r$ for simplicity. This is not very consistent, but a ring element is a ring element, whether the ring is $R$ or $R/I.$

 

[elias001]

@fresh_42 In the case of quotient topological spaces, what is the equivalent of coset elements?

 

[fresh_42]

@fresh_42 In the case of quotient topological spaces, what is the equivalent of coset elements?

In topology, everything spins around the definition of open sets. If we want to consider a quotient topology, we will need two ingredients: a surjective function $q\, : \,X\longrightarrow Y$ on the quotient space, and a definition that makes this function continuous. The quotient topology is defined by saying which sets in $Y$ are open, and these sets are exactly those that make the surjective function $q$ continuous, i.e.
$$
U\subseteq Y \text{ is open iff } q^{-1}(U)\subseteq X \text{ is open}.
$$
The problem in answering your question is that we do not have a general description of a surjective function between two topological spaces. This is fundamentally different from vector spaces or groups, where we get those equivalence classes from the function itself. But how do you describe a surjective function between arbitrary topological spaces? The best we can achieve is defining an equivalence relation on $X\times X$ and writing $Y=X/\sim.$ This reduces the problem to the definition of these equivalence classes, which vary from case to case. The answer to your question is now $x\sim \tilde{x} \Longleftrightarrow q(x)=q(\tilde{x}).$

Wikipedia has some examples: https://en.wikipedia.org/wiki/Quotient_space_(topology)#Examples