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Quotient topology seems to satisfy the universal property for coproducts at first glance. However, at second glance they seem to fail to fit into that frame in general since not every map passes to the quotient. So, do I miss some argument or is it simply that the quotient topology belongs to another categorical notion?

To be more precise:

The quotient topology satisfies a universal property in the sense:
A surjective map from some initial topological space induces a unique topology on its codomain s.t. whenever theres a map from the quotient space to some arbitrary topological space then it is continuous iff its composition with the quotient map is continuous. ... But this is somehow stated not in the right direction as it is given for coproducts, i.e. given a map rather from the initial topological space there exists a unique continuous map from the quotient space. This fails to exist iff the map is not constant on fibres.

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  • $\begingroup$ could you please be more precise? $\endgroup$ – user126154 Mar 21 '14 at 18:23
  • $\begingroup$ The disjoint union is the coproduct in the category of topological spaces. What do you mean by the 'quotient topology' of some number of spaces? $\endgroup$ – Ian Coley Mar 21 '14 at 18:27
  • $\begingroup$ I don't mean a number of spaces just a single one ...so my question seems to be: Is the disjoint union the same as the quotient topology for a single space? (not sosuper sure wether I got everything properly) $\endgroup$ – C-Star-W-Star Mar 21 '14 at 18:34
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    $\begingroup$ The general construction is the "final topology". The most prominent occurrences of final topologies are disjoint sums (coproducts in the category of topological spaces with continuous maps as morphisms) and quotient topologies. $\endgroup$ – Daniel Fischer Mar 21 '14 at 18:46
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A quotient topology is best thought of as a coequalizer. In category theory, a coequalizer is the colimit of a diagram:

$X \rightrightarrows Y$

In other words, if I have two maps $f,g:X \rightarrow Y$, the coequalizer is a map $p:Y \rightarrow Z$ which is universal with respect to the requirement that $p \circ f = p \circ g$. If you think of the objects as sets, this is basically saying that we identify the images $f(x)$ and $g(x)$ in $Y$.

To explicitly connect this to a quotient map, here's a construction: Suppose $p:X \rightarrow Z$ is a surjective function. Let $U$ be the subset of $X \times X$ where $(x,y) \in U$ if and only if $p(x) = p(y)$. Give $U$ the subspace topology. Then there are two natural maps $\pi_1,\pi_2:U \rightarrow X$ (projection onto each coordinate). The claim is that, if we give $Z$ the quotient topology, $p$ becomes the colimit of the diagram:

$U \rightrightarrows X$

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  • $\begingroup$ Some worthwhile vocabulary: when $\pi_1,\pi_2\colon U\to X$ is the pullback of $p,p\colon X\to Z$, the triple $(U,\pi_1,\pi_2)$ is called the kernel pair of $p$. Kernel pairs always have the extra structure of being (internal) equivalence relations/(internal) groupoids. For the kernel pair of $p$ two (generalized) elements are equivalent if they are in the same fiber of $p$. An equivalence relation is called effective if it has a coequalizer, i.e. an object of (internal) equivalence classes. The quotient space of $p$ is such a coequalizer and thus the space of fibers of $p$. $\endgroup$ – Vladimir Sotirov Mar 22 '14 at 0:55
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    $\begingroup$ I think that the notion of coequalizer don't permit to explain the property asked in the question, namely: a map from the quotient space to some arbitrary topological space then it is continuous iff its composition with the quotient map is continuous. This property is generalized by the costruction of final morphism in a concrete category. $\endgroup$ – Fabio Lucchini Mar 22 '14 at 9:22
  • $\begingroup$ @FabioLucchini It does in fact explain it...let $F$ be the forgetful functor, and let $f, g$ be set maps such that $f = g \circ F(p)$, where $p$ is the quotient map. We want to show that exists $\tilde{g}$ such that $g=F(\tilde{g})$ iff exists $\tilde{f}$ such that $\tilde{f}=F(f)$. The "only if" direction is trivial, since then $f = F(\tilde{g} \circ p)$. For the "if" direction, by property of coequalizer we know that if $\tilde{f} \circ \pi_1 = \tilde{f} \circ \pi_2$ then there exists a unique $\tilde{g}$ such that $\tilde{f} = \tilde{g}\circ p$. $\endgroup$ – Eric Auld Dec 6 '14 at 8:23
  • $\begingroup$ @FabioLucchini Now $F(\tilde{f}) = F(\tilde{g}) \circ F(p) = g \circ F(p)$, but $F(p)$ is an epimorphism in $\mathsf{Set}$, so $g = F(\tilde{g})$. $\endgroup$ – Eric Auld Dec 6 '14 at 8:24
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Let $X$, $Y$ be two set, $\tau$ be a topology on $X$ and $f:X\to Y$ be a function. Then $\sigma=\{V\subseteq Y:f^{-1}(V)\in\tau\}$ is the unique topology on $Y$ which satisfy the following property: For each topological space $(Z,\nu)$ and function $g:Y\to Z$, $g:(Y,\sigma)\to (Z,\nu)$ is continuos if and only if $g\circ f:(X,\tau)\to (Z,\nu)$ is continuous.

If $R$ is an equivalence relation on $X$ and $Y=X\diagup R$ and $f$ is the canonical projection, then $\sigma$ is called quotient topology.

In the language of category theory, let $\mathscr C$ be a concrete category over $\mathscr X$ with forgetful functor $|\cdot|:\mathscr C\to\mathscr X$. A morphism $f:X\to Y$ is final when ever for any object $Z$ of $\mathscr C$ and morphism $g:|Y|\to |Z|$ if $|f||g|$ lift to a morphism in $\mathscr C$ then $g$ lift to a morphism in $\mathscr C$.

If you consider the category $\mathfrak{Top}$ of topological spaces as concrete category over the category $\mathfrak{Set}$ of sets, the $f:(X,\tau)\to (Y,\sigma)$ defined above is final.

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No it is not quite the same since a quotient is defined by a characteristic property while a coproduction is defined by a universal property (see discussion/answer on Proof: Categorical Product = Topological Product)

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