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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 24th 2010
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexAdded Baez-Hoffnung reference to [[quasitopos]], as well as the result that [[concrete sheaves]] on a [[concrete site]] for a quasitopos

   Added Baez-Hoffnung reference to quasitopos, as well as the result that concrete sheaves on a concrete site for a quasitopos

2. • CommentRowNumber2.
   • CommentAuthorRodMcGuire
   • CommentTimeSep 6th 2017
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   Author: RodMcGuire
   Format: MarkdownItexI've added some tags to [[quasitopos##exampsep]] as well as links to some of the separated categories.

   I’ve added some tags to quasitopos##exampsep as well as links to some of the separated categories.

3. • CommentRowNumber3.
   • CommentAuthorRodMcGuire
   • CommentTimeSep 13th 2017
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   Author: RodMcGuire
   Format: MarkdownItexI've added to [[quasitopos#idea]] a bit more summary information and a terminology note on [[strong monomorphism]] vs [[regular monomorphism]].

   I’ve added to quasitopos#idea a bit more summary information and a terminology note on strong monomorphism vs regular monomorphism.

4. • CommentRowNumber4.
   • CommentAuthorRodMcGuire
   • CommentTimeSep 13th 2017
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   Author: RodMcGuire
   Format: MarkdownItexI'm confused A [[quasitopos]] can be [[balanced category|unbalenced]]. [[strong monomorphism]] says > * Every [[regular monomorphism]] is strong. [[regular epimorphism]] says > this implies that a regular epimorphism which is also a [[monomorphism]] must in fact be an [[isomorphism]]. [[regular monomorphism]] doesn't say the corresponding statement but says to look at [[regular epimorphism]] for dual statements. I'm not sure which pages need fixing if they do. I believe a correct statement is that in a [[quasitopos]] a [[strong monomorphism]] and a [[regular monomorphism]] are the same.

   I’m confused

   A quasitopos can be unbalenced.

   strong monomorphism says

   • Every regular monomorphism is strong.

   regular epimorphism says

   this implies that a regular epimorphism which is also a monomorphism must in fact be an isomorphism.

   regular monomorphism doesn’t say the corresponding statement but says to look at regular epimorphism for dual statements.

   I’m not sure which pages need fixing if they do. I believe a correct statement is that in a quasitopos a strong monomorphism and a regular monomorphism are the same.

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 13th 2017
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   Author: Todd_Trimble
   Format: MarkdownItexWhat are you confused by? All those look like correct statements to me. In case it needs saying, one of the main points here is that in a quasitopos, monomomorphisms need not be regular. Whereas in a topos, all monomorphisms *are* regular. Let me prove that last fact, because it's easy. According to the definition of subobject classifier in a topos, any monomorphism $i: A \to B$ is obtained by pulling back a universal monomorphism $t: 1 \to \Omega$ along a unique map $\chi_i: B \to \Omega$ called the *classifying map* of $i$: $$\array{ A & \stackrel{!_A}{\to} & 1 \\ \mathllap{i} \downarrow & (pb) & \downarrow \mathrlap{t} \\ B & \underset{\chi_i}{\to} & \Omega }$$ But this is equivalent to saying that $i: A \to B$ is an equalizer of the pair of maps $\chi_i : B \to \Omega$ and $B \stackrel{!_B}{\to} 1 \stackrel{t}{\to} \Omega$. Hence every monomorphism $i$ is an equalizer of some pair of arrows, i.e., every monomorphism is a regular monomorphism. The same argument proves generally that given some class of monomorphisms $\mathcal{M}$ closed under composition, isomorphisms, pullbacks, etc., if we have an $\mathcal{M}$-subobject classifier, i.e., every $\mathcal{M}$-subobject $i$ of $B$ is obtained by pulling back a universal $\mathcal{M}$-map $t: 1 \to \Omega$ along a unique morphism $\chi_i: B \to \Omega$, then every $\mathcal{M}$-monomorphism is regular. The converse of this statement for $\mathcal{M} =$ regular monos is the subobject classifier axiom for a quasitopos. A corollary of the argument is that a topos is balanced, i.e., that every epic monomorphism is an isomorphism. This is because epic monomorphisms are epic regular monomorphisms, i.e., epic equalizers, and such are isomorphisms. For if $e: A \to B$ is epic and is the equalizer of $f, g: B \rightrightarrows X$, then from the equalizing equation $f e = g e$ and $e$'s being epic, we deduce $f = g$. But the equalizer of an identical pair $f = g$ must be (isomorphic to) $1_B: B \to B$ -- that is to say, the equalizer must be an isomorphism. But of course this argument does not extend to show that a quasitopos is balanced, for the simple reason that monomorphisms in a quasitopos need not be regular. An intuitive class of examples to keep in mind for this are the various topological quasitoposes one often sees, such as the category of [[pseudotopological spaces]] or [[equilogical spaces]]. Just as in $Top$ a monomorphism $i: A \to B$ is regular if and only if the factoring $A \to i(A)$ through the image with the subspace topology is an isomorphism, so it is in $PseudoTop$ that regular monomomorphisms are essentially the same as subspaces, so that (for example) the identity map $id: \mathbb{R}_{disc} \to \mathbb{R}$ (from $\mathbb{R}$ with the discrete topology to $\mathbb{R}$ with the standard topology) is an example of a monomorphism that is not regular in $PseudoTop$.

   What are you confused by? All those look like correct statements to me.

   In case it needs saying, one of the main points here is that in a quasitopos, monomomorphisms need not be regular. Whereas in a topos, all monomorphisms are regular.

   Let me prove that last fact, because it’s easy. According to the definition of subobject classifier in a topos, any monomorphism $i: A \to B$ is obtained by pulling back a universal monomorphism $t: 1 \to \Omega$ along a unique map $\chi_i: B \to \Omega$ called the classifying map of $i$:

   $$\array{ A & \stackrel{!_A}{\to} & 1 \\ \mathllap{i} \downarrow & (pb) & \downarrow \mathrlap{t} \\ B & \underset{\chi_i}{\to} & \Omega }$$

   But this is equivalent to saying that $i: A \to B$ is an equalizer of the pair of maps $\chi_i : B \to \Omega$ and $B \stackrel{!_B}{\to} 1 \stackrel{t}{\to} \Omega$. Hence every monomorphism $i$ is an equalizer of some pair of arrows, i.e., every monomorphism is a regular monomorphism.

   The same argument proves generally that given some class of monomorphisms $\mathcal{M}$ closed under composition, isomorphisms, pullbacks, etc., if we have an $\mathcal{M}$-subobject classifier, i.e., every $\mathcal{M}$-subobject $i$ of $B$ is obtained by pulling back a universal $\mathcal{M}$-map $t: 1 \to \Omega$ along a unique morphism $\chi_i: B \to \Omega$, then every $\mathcal{M}$-monomorphism is regular. The converse of this statement for $\mathcal{M} =$ regular monos is the subobject classifier axiom for a quasitopos.

   A corollary of the argument is that a topos is balanced, i.e., that every epic monomorphism is an isomorphism. This is because epic monomorphisms are epic regular monomorphisms, i.e., epic equalizers, and such are isomorphisms. For if $e: A \to B$ is epic and is the equalizer of $f, g: B \rightrightarrows X$, then from the equalizing equation $f e = g e$ and $e$’s being epic, we deduce $f = g$. But the equalizer of an identical pair $f = g$ must be (isomorphic to) $1_B: B \to B$ – that is to say, the equalizer must be an isomorphism.

   But of course this argument does not extend to show that a quasitopos is balanced, for the simple reason that monomorphisms in a quasitopos need not be regular. An intuitive class of examples to keep in mind for this are the various topological quasitoposes one often sees, such as the category of pseudotopological spaces or equilogical spaces. Just as in $Top$ a monomorphism $i: A \to B$ is regular if and only if the factoring $A \to i(A)$ through the image with the subspace topology is an isomorphism, so it is in $PseudoTop$ that regular monomomorphisms are essentially the same as subspaces, so that (for example) the identity map $id: \mathbb{R}_{disc} \to \mathbb{R}$ (from $\mathbb{R}$ with the discrete topology to $\mathbb{R}$ with the standard topology) is an example of a monomorphism that is not regular in $PseudoTop$.