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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.2012-06-14
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    @ArturoMagidin Thank you. But I'm not clear with 'the quotient map $G\to G/\sim$ is not the universal on the "right side" to be translated by the free object functor', could you please give me some more details?2012-06-14
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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)2012-06-14
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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 sets2012-06-14
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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.2012-06-14
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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.2012-06-14
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    Thank you very much. Let me sleep on it.2012-06-14
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    Here is another way to motivate that equivalence relations are good enough for topological spaces: a congruence relation on an (universal) algebra $A$ is the same as a subalgebra $R\subseteq A\times A$ which at the same time also an equivalence relation, i.e. an equivalence relation *in* the respective category of algebras. The corresponding notion for a topological space $X$ is then just a subspace $R\subseteq X\times X$ which is also an equivalence relation.2012-06-22
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    Exactly. By the way, subspace in topology is defined in different way too. It doesn't need to be openset-preserving.2012-06-25

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You almost got it with the first definition. The problem is you assumed congruence can be defined globally, for the whole topological space. The correct approach is to say an equivalence relation $\sim$ is a congruence relation with respect to an open set $O \in \mathscr{T}$ if $$\forall x,y(Ox\land x \sim y \to Oy)$$

Note that, if $\sim$ is a congruence relation w.r.t. the unary relation $O$ on $X$, then we can define a unary relation $O'$ on $X/{\sim}$ by $$O'[x] \iff Ox$$

The quotient space is, then, $(X/{\sim}, \mathscr{T}')$, where $$\mathscr{T}' = \{O': O\in\mathscr{T}\text{ and } \sim \text{ is a congruence relation w.r.t. } O\}$$

Final remark: we can generalize the above definition for quotients of arbitrary structures. We throw away all relations (and operations) for which $\sim$ is not a congruence relation. We then define the quotient relations (and operations) in the natural way. This generalization works for topological quotients, whereas the usual definition doesn't. However, such generalization doesn't work for quotient graphs, whereas the usual definition does. So, as far as model theory is concerned, topological quotients and graphical quotients are different concepts.