# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorDavidRoberts
• CommentTimeAug 31st 2010

Added a bit to skeleton about skeletons of internal categories

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeSep 2nd 2010
• (edited Sep 2nd 2010)

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 ?

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeSep 2nd 2010
Quite right, Zoran. Well spotted.
• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeSep 2nd 2010
• (edited Sep 2nd 2010)

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).

• CommentRowNumber5.
• CommentAuthorzskoda
• CommentTimeSep 2nd 2010
OK, I will go to incorporate the remark into the entry, you can feel free to improve the notation etc. Check it.
• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeSep 2nd 2010

OK, I did.