pullbackPullback22003-08-08 12:20:542005-10-26 16:08:29Definition% this is the default PlanetMath preamble. as your knowledge
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\newcommand*{\abs}[1]{| #1 |}{\bf Definition}
Suppose $X,Y,Z$ are sets, and we have maps
\begin{eqnarray*}
f\colon Y&\to& Z, \\
\Phi\colon X&\to& Y.
\end{eqnarray*}
Then the {\bf pullback} of $f$ under $\Phi$ is the mapping
\begin{eqnarray*}
\Phi^\ast f\colon X &\to& Z, \\
x&\mapsto& (f\circ\Phi)(x).
\end{eqnarray*}
Let us denote by $M(X,Y)$ the set of all mappings $f\colon X\to Y$.
We then see that $\Phi^\ast$ is a mapping $M(Y,Z)\to M(X,Z)$.
In other words, $\Phi^\ast$ pulls back the set where $f$ is
defined on from $Y$ to $X$. This is illustrated in the below diagram.
$$
\xymatrix{
X \ar[r]^\Phi\ar[dr]_{\Phi^\ast f} & Y \ar[d]_{f} \\
& Z
}
$$
\subsubsection{Properties}
\begin{enumerate}
\item For any set $X$,
$(\operatorname{id}_X)^\ast = \operatorname{id}_{M(X,X)}$.
\item Suppose we have maps
\begin{eqnarray*}
\Phi\colon X&\to& Y, \\
\Psi\colon Y&\to& Z
\end{eqnarray*}
between sets $X,Y,Z$. Then
$$ (\Psi\circ \Phi)^\ast = \Phi^\ast \circ \Psi^\ast.$$
\item If $\Phi\colon X\to Y$ is a bijection, then
$\Phi^\ast$ is a bijection and
$$
\big(\Phi^\ast\big)^{-1} = \big(\Phi^{-1}\big)^\ast.
$$
\item Suppose $X,Y$ are sets with $X\subset Y$.
Then we have the inclusion map $\iota:X\hookrightarrow Y$, and
for any $f\colon Y\to Z$, we have
$$
\iota^\ast f = f|_X,
$$
where $f|_X$ is the \PMlinkname{restriction}{RestrictionOfAFunction} of $f$ to $X$.
\end{enumerate}