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2-category 2-category-theory abelian-categories adjoint algebra algebraic algebraic-geometry algebraic-topology analysis analytic-geometry arithmetic arithmetic-geometry book bundles calculus categorical categories category category-theory chern-weil-theory cohesion cohesive-homotopy-type-theory cohomology colimits combinatorics complex complex-geometry computable-mathematics computer-science constructive cosmology deformation-theory descent diagrams differential differential-cohomology differential-equations differential-geometry digraphs duality elliptic-cohomology enriched fibration foundation foundations functional-analysis functor gauge-theory gebra geometric-quantization geometry graph graphs gravity grothendieck group group-theory harmonic-analysis higher higher-algebra higher-category-theory higher-differential-geometry higher-geometry higher-lie-theory higher-topos-theory homological homological-algebra homotopy homotopy-theory homotopy-type-theory index-theory integration integration-theory k-theory lie-theory limits linear linear-algebra locale localization logic mathematics measure-theory modal modal-logic model model-category-theory monad monads monoidal monoidal-category-theory morphism motives motivic-cohomology nforum nlab noncommutative noncommutative-geometry number-theory of operads operator operator-algebra order-theory pages pasting philosophy physics pro-object probability probability-theory quantization quantum quantum-field quantum-field-theory quantum-mechanics quantum-physics quantum-theory question representation representation-theory riemannian-geometry scheme schemes set set-theory sheaf sheaves simplicial space spin-geometry stable-homotopy-theory stack string string-theory superalgebra supergeometry svg symplectic-geometry synthetic-differential-geometry terminology theory topology topos topos-theory tqft type type-theory universal variational-calculus

1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeAug 31st 2010
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexAdded a bit to [[skeleton]] about skeletons of internal categories

   Added a bit to skeleton about skeletons of internal categories

2. • CommentRowNumber2.
   • CommentAuthorzskoda
   • CommentTimeSep 2nd 2010
   • (edited Sep 2nd 2010)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexEntry skeleton says: "If the axiom of choice holds, then every category has a skeleton: simply choose one object in each isomorphism class." When the construction *is* simple, the proof is a tiny bit more subtle, namely one has to show that the inclusion $in:sk(C)\to C$ is really an equivalence. For the weak inverse one forms a functor $-':x\mapsto x'$ as follows. For every object $x$ one has chosen already the unique object $x'$ in $sk(X)$ isomorphic to $x$, but one also needs to make *a choice of isomorphism* $i_x:x\to x'$ for every $x$. This enables to conjugate between $C(x,y)$ and $C(x',y')$ by $$ (x\stackrel{f}\to y)\mapsto (x'\stackrel{i_x^{-1}}\to x\stackrel{f}\to y\stackrel{i_y}\to y'). $$ This correspondence makes $-'$ a functor, that is the rule for morphisms $f' := i_y\circ f\circ i_{x}^{-1}$ is functorial. Let us show that $-'$ is a weak inverse of $in$. In one direction, $(in_{y})' = y$ for $y\in sk(C)$; in another direction notice that $i^{-1}_{in_{x'}}:in_{x'}\cong x$ for $x\in C$ is an isomorphism. It suffices to show that these isomorphisms form a *natural* isomorphism $i^{-1}_{in}:in_{-'}\to id_C$; the naturality diagram is commutative precisely because of the conjugation formula for the functor $-'$ for morphisms. Right ?

   Entry skeleton says: “If the axiom of choice holds, then every category has a skeleton: simply choose one object in each isomorphism class.”

   When the construction is simple, the proof is a tiny bit more subtle, namely one has to show that the inclusion $in:sk(C)\to C$ is really an equivalence. For the weak inverse one forms a functor $-':x\mapsto x'$ as follows. For every object $x$ one has chosen already the unique object $x'$ in $sk(X)$ isomorphic to $x$, but one also needs to make a choice of isomorphism $i_x:x\to x'$ for every $x$. This enables to conjugate between $C(x,y)$ and $C(x',y')$ by

   $$(x\stackrel{f}\to y)\mapsto (x'\stackrel{i_x^{-1}}\to x\stackrel{f}\to y\stackrel{i_y}\to y').$$

   This correspondence makes $-'$ a functor, that is the rule for morphisms $f' := i_y\circ f\circ i_{x}^{-1}$ is functorial. Let us show that $-'$ is a weak inverse of $in$. In one direction, $(in_{y})' = y$ for $y\in sk(C)$; in another direction notice that $i^{-1}_{in_{x'}}:in_{x'}\cong x$ for $x\in C$ is an isomorphism. It suffices to show that these isomorphisms form a natural isomorphism $i^{-1}_{in}:in_{-'}\to id_C$; the naturality diagram is commutative precisely because of the conjugation formula for the functor $-'$ for morphisms.

   Right ?

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 2nd 2010
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   Author: Todd_Trimble
   Format: TextQuite right, Zoran. Well spotted.

   Quite right, Zoran. Well spotted.
4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeSep 2nd 2010
   • (edited Sep 2nd 2010)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexRight, it should say "simply choose one object in each isomorphism class and one isomorphism to that object from each other object in that class" (or something like that).

   Right, it should say “simply choose one object in each isomorphism class and one isomorphism to that object from each other object in that class” (or something like that).

5. • CommentRowNumber5.
   • CommentAuthorzskoda
   • CommentTimeSep 2nd 2010
   • PermaLink
   Author: zskoda
   Format: TextOK, I will go to incorporate the remark into the [[skeleton|entry]], you can feel free to improve the notation etc. Check it.

   OK, I will go to incorporate the remark into the entry, you can feel free to improve the notation etc. Check it.
6. • CommentRowNumber6.
   • CommentAuthorTobyBartels
   • CommentTimeSep 2nd 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexOK, I did.

   OK, I did.