essentially surjectiveEssentiallySurjective2005-05-22 18:03:522007-11-22 01:41:46Definitionisomorphism-dense subcategory\usepackage{amssymb,amscd}
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Let $\mathcal{C}$ and $\mathcal{D}$ be categories. A functor $F\colon \mathcal{C}\to \mathcal{D}$ is \emph{essentially surjective} if for any object $A\in\mathcal{OB}(\mathcal{D})$, there exists an object $X\in\mathcal{OB}(\mathcal{C})$, such that $F(X)\cong A$. That is, there are morphisms (in $D$) $f \colon F(X)\to A$ and $g\colon A\to F(X)$ such that $fg=1_A$ and $gf=1_{F(X)}$.
\textbf{Remarks.}
\begin{itemize}
\item Clearly, if $F$ is surjective, it is essentially surjective. But the reverse is not true.
\item A functor is an \PMlinkname{equivalence}{EquivalenceOfCategories} iff it is \PMlinkname{full}{FullFunctor}, \PMlinkname{faithful}{FaithfulFunctor} and essentially surjective.
\item \textbf{isomorphism-dense subcategory}. A full subcategory $\mathcal{S}$ of a category $\mathcal{C}$ is said to be \emph{isomorphism-dense in} $\mathcal{C}$, if the inclusion functor $\mathcal{S}\hookrightarrow \mathcal{C}$ is essentially surjective. Since $\mathcal{S}$ is full, the inclusion functor is full and faithful. As a result, $\mathcal{S}$ is isomorphism-dense if the inclusion functor is an equivalence.
\end{itemize}