identification topologyIdentificationTopology2004-10-05 16:01:472008-01-22 00:36:37Definition% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\real}{\mathbb{R}}Let $f$ be a function from a topological space $X$ to a set $Y$. The \emph{identification topology} on $Y$ with respect to $f$ is defined to be the finest topology on $Y$ such that the function $f$ is continuous.
\begin{thm} Let $f:X\to Y$ be defined as above. The following are equivalent:
\begin{enumerate}
\item $\mathcal{T}$ is the identification topology on $Y$.
\item $U\subseteq Y$ is open under $\mathcal{T}$ iff $f^{-1}(U)$ is open in $X$.
\end{enumerate}
\end{thm}
\begin{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 $\mathcal{T}'$.
\end{proof}
\textbf{Remarks}.
\begin{itemize}
\item
$\mathcal{S}=\lbrace f(V)\mid V\mbox{ is open in }X\rbrace$ is a subbasis for $f(X)$, using the subspace topology on $f(X)$ of the identification topology on $Y$.
\item
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 \emph{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 \emph{final topology}.
\item
The dual concept of this is the initial topology.
\item
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.
\end{itemize}