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identification topology
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(Definition)
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Let $f$ be a function from a topological space $X$ to a set $Y$ . The identification topology on $Y$ with respect to $f$ is defined to be the finest topology on $Y$ such that the function $f$ is continuous.
Theorem 1 Let $f:X\to Y$ be defined as above. The following are equivalent:
- $\mathcal{T}$ is the identification topology on $Y$ .
- $U\subseteq Y$ is open under $\mathcal{T}$ iff $f^{-1}(U)$ is open in $X$ .
Proof. ( $1.\Rightarrow 2.$ ) If $U$ is open under $\mathcal{T}$ , then $f^{-1}(U)$ is open in $X$ as $f$ is continuous under $\mathcal{T}$ . Now, suppose $U$ is not open under $\mathcal{T}$ and $f^{-1}(U)$ is open in $X$ . Let $\mathcal{B}$ be a subbase of $\mathcal{T}$ . Define $\mathcal{B}':=\mathcal{B}\cup \lbrace U\rbrace$ . Then the topology $\mathcal{T}'$ generated by $\mathcal{B}'$ is a strictly finer topology than $\mathcal{T}$ making $f$ continuous, a contradiction.
($2.\Rightarrow 1.$ ) Let $\mathcal{T}$ be the topology defined by 2. Then $f$ is continuous. Suppose $\mathcal{T}'$ is another topology on $Y$ making $f$ continuous. Let $U$ be $\mathcal{T}'$ -open. Then $f^{-1}(U)$ is open in $X$ , which implies $U$ is $\mathcal{T}$ -open. Thus $\mathcal{T}'\subseteq \mathcal{T}$ and $\mathcal{T}$ is finer than $\mathcal{T}'$ . 
Remarks.
- $\mathcal{S}=\lbrace f(V)\mid V\mbox{ is open in }X\rbrace$ is a subbasis for $f(X)$ , using the subspace topology on $f(X)$ of the identification topology on $Y$ .
- More generally, let $X_i$ be a family of topological spaces and $f_i:X_i\to Y$ be a family of functions from $X_i$ into $Y$ . The identification topology on $Y$ with respect to the family $f_i$ is the finest topology on $Y$ making each $f_i$ a continuous function. In literature, this topology is also called the final topology.
- The dual concept of this is the initial topology.
- Let $f:X\to Y$ be defined as above. Define binary relation $\sim$ on $X$ so that $x\sim y$ iff $f(x)=f(y)$ . Clearly $\sim$ is an equivalence relation. Let $X^*$ be the quotient $X/\sim$ . Then $f$ induces an injective map $f^*:X^*\to Y$ given by $f^*([x])=f(x)$ . Let $Y$ be given the identification topology and $X^*$ the quotient topology (induced by $\sim$ ), then $f^*$ is continuous. Indeed, for if $V\subseteq Y$ is open, then $f^{-1}(V)$ is open in $X$ . But then $f^{-1}(V)=\bigcup f^{* -1}(V)$ , which implies $f^{* -1}(V)$ is open in $X^*$ . Furthermore, the argument is reversible, so that if $U$ is open in $X^*$ , then so is $f^*(U)$ open in $Y$ . Finally, if $f$ is surjective, so is $f^*$ , so that $f^*$ is a homeomorphism.
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"identification topology" is owned by rspuzio. [ full author list (2) | owner history (2) ]
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Cross-references: homeomorphism, surjective, argument, induced, quotient topology, injective, induces, quotient, equivalence relation, binary relation, initial topology, subspace topology, subbasis, finer, implies, contradiction, strictly finer, generated by, iff, open, the following are equivalent, continuous, topological space, function
This is version 8 of identification topology, born on 2004-10-05, modified 2008-01-22.
Object id is 6299, canonical name is IdentificationTopology.
Accessed 2402 times total.
Classification:
| AMS MSC: | 54A99 (General topology :: Generalities :: Miscellaneous) |
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Pending Errata and Addenda
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