# 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.

1. 1.

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

2. 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.

Title identification topology IdentificationTopology 2013-03-22 14:41:26 2013-03-22 14:41:26 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Definition msc 54A99 final topology InitialTopology