# The Stacks Project

## Tag 0063

Definition 5.19.3. Let $f : X \to Y$ be a continuous map of topological spaces.

1. We say that specializations lift along $f$ or that $f$ is specializing if given $y' \leadsto y$ in $Y$ and any $x'\in X$ with $f(x') = y'$ there exists a specialization $x' \leadsto x$ of $x'$ in $X$ such that $f(x) = y$.
2. We say that generalizations lift along $f$ or that $f$ is generalizing if given $y' \leadsto y$ in $Y$ and any $x\in X$ with $f(x) = y$ there exists a generalization $x' \leadsto x$ of $x$ in $X$ such that $f(x') = y'$.

The code snippet corresponding to this tag is a part of the file topology.tex and is located in lines 3246–3259 (see updates for more information).

\begin{definition}
\label{definition-lift-specializations}
Let $f : X \to Y$ be a continuous map of topological spaces.
\begin{enumerate}
\item We say that {\it specializations lift along $f$} or that $f$ is
{\it specializing} if given $y' \leadsto y$ in $Y$ and any $x'\in X$ with
$f(x') = y'$ there exists a specialization $x' \leadsto x$ of $x'$ in $X$ such
that $f(x) = y$.
\item We say that {\it generalizations lift along $f$} or that $f$ is
{\it generalizing} if given $y' \leadsto y$ in $Y$ and any $x\in X$ with
$f(x) = y$ there exists a generalization $x' \leadsto x$ of $x$ in $X$ such
that $f(x') = y'$.
\end{enumerate}
\end{definition}

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