396: Universal Property of Quotient Map

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A description/proof of universal property of quotient map

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of quotient map.
• The reader knows a definition of quotient topology.
• The reader admits the proposition that for any map, the composition of the map after any preimage is identical if and only if the argument set is a subset of the map image.

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Target Context

• The reader will have a description and a proof of the universal property of quotient map: any surjection between topological spaces is a quotient map if and only if any additional map from the codomain of the original map to any additional topological space is continuous if and only if the composition of the additional map after the original map is continuous.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any topological spaces, $T_1,T_2$, and any surjection, $f_1:T_1\to T_2$, $f_1$ is a quotient map if and only if for any topological space, $T_3$, and any map, $f_2:T_2\to T_3$, $f_2$ is continuous if and only if $f_2\circ f_1:T_1\to T_3$ is continuous.

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2: Proof

Suppose that $f_1$ is a quotient map. Suppose $f_2$ is continuous. Then, $f_2\circ f_1$ is continuous as a compound of continuous maps. Suppose $f_2\circ f_1$ is continuous. Then, for any open set, $U\subseteq T_3$, ${f_2\circ f_1}^{-1}(U)={f_1}^{-1}\circ {f_2}^{-1}(U)={f_1}^{-1}({f_2}^{-1}(U))$ is open. By the definition of quotient map, ${f_2}^{-1}(U)$ is open, so, $f_2$ is continuous.

Suppose that for any topological space, $T_3$, and any map, $f_2:T_2\to T_3$, $f_2$ is continuous if and only if $f_2\circ f_1:T_1\to T_3$ is continuous. Let us take $T_3=T_2$ and $f_2:T_2\to T_2$ as the identity map, continuous. So, $f_2\circ f_1=f_1:T_1\to T_2$ is continuous. Let us take $T_3:=T_1/f_1$, which is the quotient space of $T_1$ such that any pair, $p_1,p_2\in T_1,f_1(p_1)=f_1(p_2)$, are identified. Let us take $f_2:f_1(p)\mapsto [p]$, which is obviously bijective. $f_2\circ f_1$, which is really the canonical map to the quotient space, is continuous, because for any open set, $U\subseteq T_3$, $(f_2\circ f_1{)}^{-1}(U)$ is open by the definition of quotient topology. So, by the supposition, $f_2$ is continuous. Now, for any subset, $S\subseteq T_2$, if ${f_1}^{-1}(S)$ is open, $f_2\circ f_1({f_1}^{-1}(S))$ is open by the definition of quotient topology, because ${f_1}^{-1}(S)=(f_2\circ f_1{)}^{-1}(f_2\circ f_1({f_1}^{-1}(S))$, which is because for any $p\in {f_1}^{-1}(S)$, $f_2\circ f_1(p)\in f_2\circ f_1({f_1}^{-1}(S))$, so, $p\in (f_2\circ f_1{)}^{-1}(f_2\circ f_1({f_1}^{-1}(S))$, and for any $p\in (f_2\circ f_1{)}^{-1}(f_2\circ f_1({f_1}^{-1}(S))$, $f_2\circ f_1(p)\in f_2\circ f_1({f_1}^{-1}(S))$, but as $f_2$ is bijective, $f_1(p)\in f_1({f_1}^{-1}(S))$, but as $f_1$ is surjective, by the proposition that for any map, the composition of the map after any preimage is identical if and only if the argument set is a subset of the map image, $f_1\circ {f_1}^{-1}(S)=S$, so, $f_1(p)\in S$. $S={f_2}^{-1}((f_2\circ f_1({f_1}^{-1}(S)))$ because $f_2$ is bijective, but as $f_2$ is continuous, $S$ is open.

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References

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