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\begin{document}

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\section*{connected object}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{compact_objects}{}\paragraph*{{Compact objects}}\label{compact_objects}

[[!include compact object - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{characterization_in_terms_of_coproducts}{Characterization in terms of coproducts}\dotfill \pageref*{characterization_in_terms_of_coproducts} \linebreak
\noindent\hyperlink{general_properties}{General properties}\dotfill \pageref*{general_properties} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A connected object is a generalisation of the concept of [[connected space]] from [[Top]] to an arbitrary [[extensive category]].

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

Let $C$ be an [[extensive category]], then:

\begin{defn}
\label{ConnectedObject}\hypertarget{ConnectedObject}{}
An [[object]] $X$ of $C$ is \textbf{connected} if the [[representable functor]]

\begin{displaymath}
hom(X, -) \colon C \to Set
\end{displaymath}
[[preserved limit|preserves]] all [[coproducts]].

\end{defn}
\begin{remark}
\label{}\hypertarget{}{}
By definition, $hom(X,-)$ preserves binary coproducts if the canonically defined morphism $hom(X,Y) + hom(X,Z) \to hom(X,Y + Z)$ is always a [[bijection]].

\end{remark}
\begin{remark}
\label{}\hypertarget{}{}
By this definition, the [[initial object]] of $C$ is \emph{not} in general connected (except for degenerate cases of $C$); it is [[too simple to be simple]]. This matches the notion that the [[empty space]] should not be considered connected, discussed at [[connected space]].

\end{remark}
\begin{remark}
\label{}\hypertarget{}{}
If $C$ is a [[infinitary extensive category]] then for $X \in Ob(C)$ to be connected it is enough to require that $hom(X,-)$ preserves \emph{binary} coproducts. This is theorem \ref{RespectForBinaryCoproductsIsSufficient} below.

\end{remark}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{characterization_in_terms_of_coproducts}{}\subsubsection*{{Characterization in terms of coproducts}}\label{characterization_in_terms_of_coproducts}

Let $C$ be an infinitary [[extensive category]], then

\begin{theorem}
\label{RespectForBinaryCoproductsIsSufficient}\hypertarget{RespectForBinaryCoproductsIsSufficient}{}
An [[object]] $X$ of $C$ is connected, def. \ref{ConnectedObject}, if and only if $\hom(X, -): C \to Set$ preserves binary coproducts.

\end{theorem}
\begin{proof}
The ``only if'' is clear, so we just prove the ``if''.

We first show that $\hom(X, -)$ preserves the [[initial object]] $0$. Indeed, if $\hom(X, -)$ preserves the binary product $X + 0 = X$, then the canonical map

\begin{displaymath}
\hom(X, X) + \hom(X, 0) \to \hom(X, X)
\end{displaymath}
is a bijection of sets, where the restriction to $\hom(X, X)$ is also a bijection of sets $id: \hom(X, X) \to \hom(X, X)$. This forces the set $\hom(X, 0)$ to be empty.

Now let $\{Y_\alpha: \alpha \in A\}$ be a set of objects of $C$. We are required to show that each map

\begin{displaymath}
f\colon X \to \sum_\alpha Y_\alpha
\end{displaymath}
factors through a unique inclusion $i_\alpha: Y_\alpha \to \sum_\alpha Y_\alpha$. By infinite extensivity, each pullback $U_\alpha \coloneqq f^\ast (Y_\alpha)$ exists and the canonical map $\sum_\alpha U_\alpha \to X$ is an isomorphism. We will treat it as the identity.

Now all we need is to prove the following.

\begin{itemize}%
\item Claim: $X = U_\alpha$ for exactly one $\alpha$. For the others, $U_\beta = 0$.

\end{itemize}
Indeed, for each $\alpha$, the identity map factors through one of the two summands in

\begin{displaymath}
id\colon X \to U_\alpha + \sum_{\beta \neq \alpha} U_\beta
\end{displaymath}
because $\hom(X, -)$ preserves binary coproducts. In others words, either $X = U_\alpha$ or $X = \sum_{\beta \neq \alpha} U_\beta$ (and the other is $0$). We cannot have $U_\alpha = 0$ for every $\alpha$, for then $X = \sum_\alpha U_\alpha$ would be $0$, contradicting the fact that $\hom(X, 0) = 0$. So $X = U_\alpha$ for at least one $\alpha$. And no more than one $\alpha$, since we have $U_\alpha \cap U_\beta = 0$ whenever $\alpha \neq \beta$.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
This proof is not [[constructive mathematics|constructive]], as we have no way to construct a particular $\alpha$ such that $X = U_\alpha$. (It is constructive if [[Markov's principle]] applies to $A$.)

\end{remark}
From the proof above, we extract a result useful in its own right, giving an alternative definition of connected object.

\begin{theorem}
\label{Scholium}\hypertarget{Scholium}{}
An object $X$ in an [[extensive category]] is connected, def. \ref{ConnectedObject}, if and only if in any [[coproduct]] decomposition $X \simeq U + V$, exactly one of $U$, $V$ is not the [[initial object]].

\end{theorem}
\begin{proof}
If $X$ is connected and $X = U + V$ is a coproduct decomposition, then the arrow $id \colon X \to U + V$, factors through one of the coproduct inclusions of $i_U, i_V \colon U, V \hookrightarrow U + V$. If it factors through say $U$, then the subobject $i_U$ is all of $X$, and $V$ is forced to be initial by disjointness of coproducts.

Turning now to the if direction, suppose $f \colon X \to Y + Z$ is a map, and put $U = f^\ast(i_Y)$, $V = f^\ast(i_Z)$. By extensivity, we have a coproduct decomposition $X = U + V$. One of $U$, $V$ is initial, say $V$, and then we have $X = U$, meaning that $f$ factors through $Y$, and uniquely so since $i_Y$ is monic. Hence $f$ belongs to (exactly) one of the two subsets $\hom(X, Y) \hookrightarrow \hom(X, Y + Z)$, $\hom(X, Z) \hookrightarrow \hom(X, Y + Z)$.

\end{proof}
\hypertarget{general_properties}{}\subsubsection*{{General properties}}\label{general_properties}

\begin{prop}
\label{}\hypertarget{}{}
A [[colimit]] of connected objects over a [[connected category|connected]] [[diagram]] is itself a connected object.

\end{prop}
\begin{proof}
Because coproducts in $Set$ commute with \emph{limits} of connected diagrams.

\end{proof}
\begin{prop}
\label{}\hypertarget{}{}
If $X \in Ob(C)$ is connected and $X \to Y$ is an [[epimorphism]], then $Y$ is connected.

\end{prop}
\begin{proof}
Certainly $Y$ is not [[initial object|initial]], because initial objects in extensive categories are [[strict initial object|strict]]. Suppose $Y = U + V$ (see theorem \ref{Scholium} above), so that we have an epimorphism $X \to U + V$. By connectedness of $X$, this epi factors through one of the summands, say $U$. But then the inclusion $i_U: U \hookrightarrow U + V$ is epic, in fact an epic equalizer of two maps $U + i_1, U + i_2: U + V \rightrightarrows U + V + V$. This means $i_U$ is an isomorphism; by disjointness of coproducts, this forces $V$ to be initial. Of course $U$ is not initial; otherwise $Y$ would be initial.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
It need not be the case that products of connected objects are connected. For example, in the topos $\mathbb{Z}$-Set, the product $\mathbb{Z} \times \mathbb{Z}$ decomposes as a countable coproduct of copies of $\mathbb{Z}$. (For more on this topic, see also [[cohesive topos]].)

\end{remark}
We do have the following partial result, generalizing the case of $Top$.

\begin{theorem}
\label{prod}\hypertarget{prod}{}
Suppose $C$ is a cocomplete $\infty$-[[extensive category]] with [[finite products]]. Assume that the [[terminal object]] is a [[separator]], and that [[Cartesian product]] functors $X \times (-) : C \to C$ preserve [[epimorphisms]]. Then a product of finitely many connected objects is itself connected.

\end{theorem}
\begin{proof}
In the first place, $1$ is connected. For suppose $1 = U + V$ where $U$ is not initial. The two coproduct inclusions $U \to U + U$ are distinct by disjointness of sums. Since $1$ is a separator, there must be a map $1 \to U$ separating these inclusions. We then conclude $U \cong 1$, and then $V \cong 0$ by disjointness of sums.

Now let $X$ and $Y$ be connected. The two inclusions $i_1, i_2: X \to X + X$ are distinct, so there exists a point $a \colon 1 \to X$ separating them. For each $y \colon 1 \to Y$, the +-shaped object $T_y = (X \times y) \cup (a \times Y)$ is connected (we define $T_y$ to be a sum of connected objects $X \times y$, $a \times Y$ amalgamated over the connected object $a \times y = 1 \times 1 = 1$, i.e., to be a connected colimit of connected objects). We have a map

\begin{displaymath}
\phi: \bigcup_{y \colon 1 \to Y} T_y \to X \times Y
\end{displaymath}
where the union is again a sum of connected objects $T_y$ amalgamated over $a \times Y$, so this union is connected. The map $\phi$ is epic, because the evident map

\begin{displaymath}
\sum_{y \colon 1 \to Y} X \times y \cong X \times \sum_{y \colon 1 \to Y} 1 \to X \times Y
\end{displaymath}
is epic (by the assumptions that $1$ is a [[separator]] and $X \times -$ preserves epis), and this map factors through $\phi$. It follows that the codomain $X \times Y$ of $\phi$ is also connected.

\end{proof}
\begin{remark}
\label{prod2}\hypertarget{prod2}{}
The same method of proof shows that for an arbitrary family of [[connected spaces]] $\{X_\alpha\}_{\alpha \in A}$, the connected component of a point $x = (x_\alpha)$ in the product space $X = \prod_{\alpha \in A} X_\alpha$ contains at least all those points which differ from $x$ in at most finitely many coordinates. However, the set of such points is dense in $\prod_{\alpha \in A} X_\alpha$, so $\prod_{\alpha \in A} X_\alpha$ must also be connected.

\end{remark}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{itemize}%
\item Connected objects in [[Top]] are precisely [[connected topological spaces]].


\item For a [[group]] $G$, connected objects in the category $G Set$ of [[permutation representation]]s are precisely the ([[inhabited set|inhabited]]) [[transitive action|transitive]] $G$-sets.


\item Objects in a [[locally connected topos]] are [[coproducts]] of connected objects.

(This includes categories such as $G Set$, [[permutation representations]] of a [[group]] $G$.)


\item A [[scheme]] is connected as a scheme (which by definition means that its underlying topological space is connected) if and only if it is connected as an object of the category of schemes. See \href{http://gausyu.github.io/2016/02/13/Connectedness-of-Schemes/}{here}.


\item [[connected graph]]



\end{itemize}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[connected topos]], [[locally connected topos]]


\item [[disjoint coproduct]]


\item [[coproduct-preserving representable]]


\item [[n-connected object in an (∞,1)-topos]]



\end{itemize}
[[!redirects connected object]] [[!redirects connected objects]]

[[!redirects connected]]



\end{document}