identification topology
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 $f\mathrm{:}X\mathrm{\to}Y$ be defined as above. The following are equivalent^{}:

1.
$\mathcal{T}$ is the identification topology on $Y$.

2.
$U\subseteq Y$ 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 ${\mathcal{B}}^{\prime}:=\mathcal{B}\cup \{U\}$. Then the topology ${\mathcal{T}}^{\prime}$ generated by ${\mathcal{B}}^{\prime}$ 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 ${\mathcal{T}}^{\prime}$ is another topology on $Y$ making $f$ continuous. Let $U$ be ${\mathcal{T}}^{\prime}$open. Then ${f}^{1}(U)$ is open in $X$, which implies $U$ is $\mathcal{T}$open. Thus ${\mathcal{T}}^{\prime}\subseteq \mathcal{T}$ and $\mathcal{T}$ is finer than ${\mathcal{T}}^{\prime}$. ∎
Remarks.

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$\mathcal{S}=\{f(V)\mid V\text{is open in}X\}$ is a subbasis for $f(X)$, using the subspace topology on $f(X)$ of the identification topology on $Y$.

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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.

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The dual concept of this is the initial topology.

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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.
Title  identification topology 

Canonical name  IdentificationTopology 
Date of creation  20130322 14:41:26 
Last modified on  20130322 14:41:26 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  11 
Author  rspuzio (6075) 
Entry type  Definition 
Classification  msc 54A99 
Synonym  final topology 
Related topic  InitialTopology 