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. $𝒯$ is the identification topology on $Y$.

2. $U\subseteq Y$ is open under $𝒯$ iff $f^{-1}(U)$ is open in $X$.

Proof.

($1.\Rightarrow 2.$) If $U$ is open under $𝒯$, then $f^{-1}(U)$ is open in $X$ as $f$ is continuous under $𝒯$. Now, suppose $U$ is not open under $𝒯$ and $f^{-1}(U)$ is open in $X$. Let $ℬ$ be a subbase of $𝒯$. Define ${ℬ}^{\prime }:=ℬ\cup \{U\}$. Then the topology ${𝒯}^{\prime }$ generated by ${ℬ}^{\prime }$ is a strictly finer topology than $𝒯$ making $f$ continuous, a contradiction.

($2.\Rightarrow 1.$) Let $𝒯$ be the topology defined by 2. Then $f$ is continuous. Suppose ${𝒯}^{\prime }$ is another topology on $Y$ making $f$ continuous. Let $U$ be ${𝒯}^{\prime }$-open. Then $f^{-1}(U)$ is open in $X$, which implies $U$ is $𝒯$-open. Thus ${𝒯}^{\prime }\subseteq 𝒯$ and $𝒯$ is finer than ${𝒯}^{\prime }$.∎

Remarks.

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

• 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.