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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 $f:X\to Y$ 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.

$\mathcal{S}=\{f(V)\mid V\mbox{ is open in }X\}$ 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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