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According to my knowledge, quotient structure is a original structure divided by a congruence. However, quotient topology space is defined this way. Quotient_topology

In this way, $\sim$ is only said be an equivalence relation, not need to be congruence. But what is the congruence on topology space?

I found two ways to define: Let $\mathfrak{X}:=(X,\mathscr{T})$ be a topology space.

  1. First if we see open sets as unary-relations on the domain, then a equivalence relation $\sim$ is a congruence if $\forall O \in \mathscr{T}\forall x,y(Ox\land x \sim y \to Oy)$ .

  2. Second if we see $\mathscr T$ as a higher-order relation with type $((0))$, then a equivalence relation $\sim$ is a congruence if $\forall E,F\subseteq X(E \in \mathscr{T}\land E \sim F \to F \in \mathscr{T})$ . Where $E \sim F$ iff $\sim[E]=\sim[F]$.

These two method are both well-defined but seem not so nice.

  1. If $\mathfrak{X}$ is $T_0$, then for each $x,y \in X$, if $x\sim y$ and $x\ne y$. Then there is an open set $O$ which contains $x$ whereas not contains $y$. But since $\sim$ is a congruence, $x\in O$ implies $y \in O$, a contradiction. That means the only possible congruence is identity.

  2. If there is an open set $O$ overcasts (in this termology $O$ overcasts $E$ iff $O \cap E \ne \emptyset$) some blocks $\{E_i\}_{i \in I}$ . then every $\bigcup_{i \in I}U_i$ must be open set too where $\forall i\in I[\emptyset \ne U_i \subseteq E_i]$. Let us consider about order topology on $\mathbb R$, since every open set is uncountable, that requires every open set must overcast uncountable many blocks. Moreover, if $E_i(i \in I)$ all contains two or more elements, then $\bigcup_{i \in I}U_i$ must not be open set where $\forall i\in I[\emptyset \ne U_i \subsetneq E_i]$ else $X\backslash \bigcup_{i \in I}U_i$ will be non-empty proper close set and there is no such non-empty clopen set in this topology space. That means $x \sim x+n$ is not a congruence.

My question is can we find a better definition?

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The quotient topology is defined to be the "universal" topology that makes the quotient map $X\to X/\sim$ continuous. The reason we can get away with this in topology and not in, say, group theory, is that the underlying set functor $\mathsf{TOP}\to\mathsf{SET}$ has both a left and a right adjoint, so that the "quotient" operation form $\mathsf{SET}$ must have an appropriate object (with that underlying set) in $\mathsf{TOP}$. In, e.g., $\mathsf{Group}$ you can't do it because the quotient map $G\to G/\sim$ is not the universal on the "right side" to be translated by the free object functor. –  Arturo Magidin Jun 14 '12 at 4:26
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A left adjoint respects colimits, a right adjoint respects limits. Limits are "right universal": you get maps into them, colimits are "left universal", you get map out of them. Thus, the underlying set of a product of groups is the product of the underlying sets (products are limits, and the underlying set functor is a right adjoint), and the free group on a disjoint union [coproduct] of sets is the free product [coproduct] of the free groups on the sets (free group is a left adjoint). (cont) –  Arturo Magidin Jun 14 '12 at 4:53
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(cont) But the underlying set of a free product is not the disjoint union of the underlying sets, and the free group of a cartesian product is not the direct product of the free groups on the factors. In topology, because the underlying set functor has both a left and a right adjoint, the underlying set of a limit is the limit of the underlying sets, and the underlying set of a colimit is the colimit of the underlying sets. So the underlying set of a product of top spaces is the product of the underlying sets, and the underlying set of the coproduct is disjoint union of the underlying sets –  Arturo Magidin Jun 14 '12 at 4:56
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(cont) The quotient map is a colimit (it is left universal). In $\mathsf{TOP}$, you can then use the fact that the underlying set functor is an adjoint on both sides to deduce that there has to be a colimit property associated to the quotient map $U(T)\to U(T)/\sim$ (where $U$ is the underlying set functor) for any equivalence relation $\sim$, which can then be "brought back" to the topological space $T$. In $\mathsf{GROUP}$ you cannot do that to $U(G)/\sim$, because after you apply the free group functor (the adjoint) you change the underlying set. –  Arturo Magidin Jun 14 '12 at 5:00
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(cont) None of this to mean that you cannot do it via a "congruence" in topology, but rather that the extra structure on the underlying set functor gives us liberties in Topology that allow us to get away with not having to do so. –  Arturo Magidin Jun 14 '12 at 5:01

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