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\begin{document}

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\section*{toposes are extensive}

\hypertarget{toposes_are_extensive}{}\section*{{Toposes are extensive}}\label{toposes_are_extensive}

\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{proofs}{Proofs}\dotfill \pageref*{proofs} \linebreak
\noindent\hyperlink{from_finitary_to_infinitary}{From finitary to infinitary}\dotfill \pageref*{from_finitary_to_infinitary} \linebreak
\noindent\hyperlink{via_pushouts_of_monos}{Via pushouts of monos}\dotfill \pageref*{via_pushouts_of_monos} \linebreak
\noindent\hyperlink{via_quasidisjointness}{Via quasi-disjointness}\dotfill \pageref*{via_quasidisjointness} \linebreak
\noindent\hyperlink{via_logical_functors}{Via logical functors}\dotfill \pageref*{via_logical_functors} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

Every [[elementary topos]] is a finitary [[extensive category]]. In fact, more is true: if an elementary topos has [[coproducts]] indexed by any [[arity class]] $\kappa$, then it is $\kappa$-extensive. In particular, a cocomplete elementary topos is infinitary extensive.

\hypertarget{proofs}{}\subsection*{{Proofs}}\label{proofs}

There are several possible proofs. In all cases, stability of coproducts under pullback is automatic since a topos is locally cartesian closed, so it suffices to consider [[disjoint coproduct|disjointness]].

\hypertarget{from_finitary_to_infinitary}{}\subsubsection*{{From finitary to infinitary}}\label{from_finitary_to_infinitary}

As shown at [[extensive category]], if we assume [[classical logic]] then $\coprod_{j\in J} A_j \cong A_1 \sqcup A_2 \sqcup \coprod_{j\neq 1,2} A_j$, so that in a finitary-extensive category, any existing coproducts are stable and disjoint. Thus, in this case it suffices to deal with finite coproducts. However, in topos theory we are often interested in toposes ``over an arbitrary base'', in which case the ambient logic is [[constructive logic]] and so this argument does not apply.

Nevertheless, we mention a couple of proofs that apparently apply only to the finitary case in addition to the fully general version.

\hypertarget{via_pushouts_of_monos}{}\subsubsection*{{Via pushouts of monos}}\label{via_pushouts_of_monos}

A proof applying to finite coproducts is given in \hyperlink{Johnstone}{Johnstone, A2.4.4}: since $0\to X$ and $0\to Y$ are monic, it suffices to show that the pushout of a mono is a mono and also a pullback.

To see this, let $f:A\rightarrowtail B$ be monic and $g:A\to C$ any morphism, and let $h:B\to P C$ be the name of $(f,g):A\to B\times C$, i.e. $(f,g)$ is the pullback $\in_C \rightarrowtail P C \times C$ along $(h,1)$. In other words, $(k,l):X\to B\times C$ factors through $(f,g)$ iff $(h k,l)$ factors through $\in_C$ (i.e. $l(x) \in h k(x)$ for all $x:X$ in the [[internal logic]]). This is in particular the case if $h k = \{\} \circ l$ (i.e. $h k(x) = \{l(x)\}$ for all $x:X$ in the internal logic), which implies that the square

\begin{displaymath}
\itexarray{A & \xrightarrow{g} & C \\ ^f\downarrow && \downarrow^{\{\}} \\ B & \xrightarrow{h} & P C }
\end{displaymath}
is a pullback.

Now given a pushout

\begin{displaymath}
\itexarray{A & \xrightarrow{g} & C \\ ^f\downarrow && \downarrow^{d} \\ B & \xrightarrow{c} & D }
\end{displaymath}
with $f$ monic, with $h$ as above we have an induced $m:D \to P C$ by the universal property of the pushout, with $m d = \{\}$ and $m c = h$. Since $\{\}$ is monic, $d$ must also be monic, and since the previous square is a pullback, so must this one be.

\hypertarget{via_quasidisjointness}{}\subsubsection*{{Via quasi-disjointness}}\label{via_quasidisjointness}

A different proof applying to finite coproducts can be obtained by combining A1.5.14 and A1.6.2 of \hyperlink{Johnstone}{Johnstone}. Define $R$ and $S$ to make the following squares pullbacks:

\begin{displaymath}
\itexarray{ R & \xrightarrow{r} & A & \xleftarrow{s} & S\\
  ^{r'}\downarrow && \downarrow^i && \downarrow\\
  A & \xrightarrow{i} & A+B & \xleftarrow{j} & B}
\end{displaymath}
Specifically, $(r,r')$ is the [[kernel pair]] of $i$. In particular, there is an induced diagonal $\triangle : A \to R$ such that $r \triangle = 1_A$. On the other hand, since pullback preserves colimits, $(r,s)$ is a coproduct diagram. Thus, the pair of morphisms $1_R:R\to R$ and $\triangle \circ s : S\to R$ factor (uniquely) through some $h:A\to R$, so that in particular $h r = 1_R$. Thus, $r:R\to A$ has both a left and a right inverse, so it is an isomorphism, and hence $i$ is monic. Dually, $j$ is monic.

Now we claim that any two morphisms $f,g:S\to X$ with domain $S$ are equal. For the pair of morphisms $R \xrightarrow{r} A \to A+X$ and $S \xrightarrow{f} X \to A+X$ factor uniquely through $A$ (which, recall, is the coproduct $R+S$). But since $r$ is an isomorphism, this factorization must be the left coprojection $A\to A+X$. Therefore, the composite $S \xrightarrow{f} X \to A+X$ is equal to the composite $S \xrightarrow{s} A \to A+X$, and similarly so is the composite $S \xrightarrow{g} X \to A+X$. Since $X\to A+X$ is monic by the previous paragraph, $f= g$.

Finally, since $0$ is a strict initial object, $0\to S$ is monic. And of course $1_S : S\to S$ is also monic, and these two monomorphisms are classified by two maps $f,g:S\to \Omega$ into the subobject classifier. By the previous paragraph, $f= g$, hence $S\cong 0$ and is initial.

\hypertarget{via_logical_functors}{}\subsubsection*{{Via logical functors}}\label{via_logical_functors}

Finally, we give a proof that applies constructively to arbitrary coproducts. This proof is due to Jibladze and can be found (in fibered-topos language) in \hyperlink{Streicher}{Streicher, Appendix A}. (Moens himself always just assumed that internal sums are stable and disjoint whereas Jibladze proved that for every fibered topos with internal sums these are actually stable and disjoint.)

Suppose $\mathcal{E}$ a topos with $J$-indexed coproducts for all $J$ belonging to some [[arity class]] $\kappa$. Then $\Delta : \mathcal{E} \to \mathcal{E}^J$ is a [[logical functor]] with a left adjoint $\coprod_J$, so (by A2.3.8 of \hyperlink{Johnstone}{Johnstone}, see also [[logical functor]]) to show that $\coprod_J$ induces equivalences $\mathcal{E}^J/\{A_j\} \to \mathcal{E}/\coprod_j A_j$ (which is one characterization of $\kappa$-extensivity) it suffices for $\coprod_J$ to be faithful. As with any adjunction, to show the left adjoint to be faithful it suffices to show that the unit $\eta : Id \to \Delta \coprod_J$ is monic, which means in this case that the injections $A_j \to \coprod_j A_j$ are all monic.

Now since $\kappa$ is an arity class, $J\times J$ also belongs to it, and hence so do the fibers of the diagonal $J\to J\times J$. Thus, $\mathcal{E}$ has coproducts indexed by the [[subsingletons]] determined by the equalities $(j=k)$ for any $j,k\in J$. Moreover, for any $j$ we can write $A_j = \coprod_{k\in J} \coprod_{(j=k)} A_{k}$, and under this identification the injection $A_j \to \coprod_{k\in J} A_{k}$ is obtained by applying the functor $\coprod_J$ to the evident family of maps $\coprod_{(j=k)} A_{k}\to A_k$. But since $\coprod_J$ is the left adjoint of a logical functor, it preserves monomorphisms (see A2.4.8 of \hyperlink{Johnstone}{Johnstone}, also [[logical functor]]). Thus, it suffices to prove that each $\coprod_{(j=k)} A_{k}\to A_k$ is monic.

Now, the functor $\coprod_{(j=k)} : \mathcal{E}^{(j=k)} \to \mathcal{E}$ has a logical right adjoint sending $B\in\mathcal{E}$ to the subsingleton-indexed family $\{B\}_{(j=k)}$, and the units of its adjunction $\{A\}_{(j=k)} \to \{ \coprod_{(j=k)} A \}_{(j=k)}$ are isomorphisms (to define an inverse, it suffices to consider the object $\coprod_{(j=k)} A$ under the assumption $j=k$, in which case this object is isomorphic to $A$). Thus, $\coprod_{(j=k)}$ is fully faithful, and in particular faithful, and since its right adjoint is also logical, it induces an equivalence $\mathcal{E}^{(j=k)} \simeq \mathcal{E}/\coprod_{(j=k)} 1$.

However, the unit of a pullback adjunction $\mathcal{E}/I \rightleftarrows \mathcal{E}$ is $A\to A\times I$, and if this is an isomorphism then in particular $I\cong I\times I$ and thus $I$ is subterminal. Moreover, in this case the counits of the adjunction are the projections $A\times I \to A$, which are pullbacks of the mono $I\to 1$ and therefore also monic. Finally, in our case where $I = \coprod_{(j=k)} 1$, the counit associated to $A_k$ is precisely the map $\coprod_{(j=k)} A_{k} \to A_k$ that we wanted to prove to be monic.

\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item [[Peter Johnstone]], [[Sketches of an Elephant]]


\item [[Thomas Streicher]], \emph{Fibered categories a la Jean Benabou}, \href{http://citeseerx.ist.psu.edu/viewdoc/citations;jsessionid=76DB3E2F57E81C74E05109B8F7E11E44?doi=10.1.1.678.919}{citeseer}



\end{itemize}


\end{document}