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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeApr 16th 2012
   • (edited Apr 16th 2012)
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexSuppose you have a category $\mathbf{C}$ and a category $\mathbf{D}$ such that $\mathbf{C}$ is a subcategory of $\mathbf{D}$ and $\mathbf{D}$ is a subcategory of $\mathbf{Set}$. Moreover the simplicial skeleton (endo)functor $\mathbf{Sk}^n$ exists in the category of simplicial $\mathbf{C}$ objects $\mathbf{sC}$ but not in the category of simplicial $\mathbf{D}$ objects $\mathbf{sD}$. Moreover suppose $X \in \mathbf{sC}$ is '$n$-skeleton' (means $\mathbf{Sk}^n(X) = X$). Then there is the adjunction $Hom_{\mathbf{sC}}(X,Y) \simeq Hom_{\mathbf{sC}}(X,\mathbf{Cosk}^n Y)$ for any $Y \in \mathbf{sC}$. Now suppose that the endofunctor $\mathbf{Cosk}^n$ still exsists in $\mathbf{sD}$ and that now $ Y \in \mathbf{sD}$ but $Y \notin \mathbf{sC}$. Does the adjunction $Hom_{\mathbf{sD}}(X,Y) \simeq Hom_{\mathbf{sD}}(X,\mathbf{Cosk}^n Y)$ still holds? _______________________________________ Something like this occurs for example when $\mathbf{C}$ is the category of vector spaces and $\mathbf{D}$ is the category of Banach manifolds and $X$ is a simplicial vector space with $\mathbf{Sk}^n X = X$ and $Y$ is a 'nonlinear' simplicial manifold.

   Suppose you have a category $C$ and a category $D$ such that $C$ is a subcategory of $D$ and $D$ is a subcategory of $\mathrm{Set}$. Moreover the simplicial skeleton (endo)functor ${\mathrm{Sk}}^n$
   exists in the category of simplicial $C$ objects $\mathrm{sC}$ but not in the category of simplicial $D$ objects $\mathrm{sD}$.

   Moreover suppose $X\in \mathrm{sC}$ is ’$n$-skeleton’ (means ${\mathrm{Sk}}^n(X)=X$). Then there is the adjunction

   ${\mathrm{Hom}}_{\mathrm{sC}}(X,Y)\simeq {\mathrm{Hom}}_{\mathrm{sC}}(X,{\mathrm{Cosk}}^{n}Y)$ for any $Y\in \mathrm{sC}$.

   Now suppose that the endofunctor ${\mathrm{Cosk}}^n$ still exsists in $\mathrm{sD}$ and that now $Y\in \mathrm{sD}$ but $Y\notin \mathrm{sC}$.

   Does the adjunction ${\mathrm{Hom}}_{\mathrm{sD}}(X,Y)\simeq {\mathrm{Hom}}_{\mathrm{sD}}(X,{\mathrm{Cosk}}^{n}Y)$ still holds?

   ---

   Something like this occurs for example when $C$ is the category of vector spaces and $D$ is the category of Banach manifolds and $X$ is a simplicial vector space with ${\mathrm{Sk}}^{n}X=X$ and $Y$ is a ’nonlinear’ simplicial manifold.

2. • CommentRowNumber2.
   • CommentAuthorMirco Richter
   • CommentTimeApr 17th 2012
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexnothing?

   nothing?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeApr 17th 2012
   • (edited Apr 17th 2012)
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   Author: Urs
   Format: MarkdownItex> Moreover suppose $X \in sC$ is '$n$-skeleton' One says $X$ is _$n$-skeletal_. I think you are after a [[relative adjoint functor]] (seel [local definition](http://ncatlab.org/nlab/show/adjoint+functor#LocalDefinition) of adjoints). To answer your question, you will need some assumption on the inclusion $C \to D$ preserving the colimits that $sk$ is built from. If they are preserved (and I take it you assume that the dual limits that $cosk$ is built from exist in $D$), then the desired relation will hold.

   Moreover suppose $X\in \mathrm{sC}$ is ’$n$-skeleton’

   One says $X$ is $n$-skeletal.

   I think you are after a relative adjoint functor (seel local definition of adjoints).

   To answer your question, you will need some assumption on the inclusion $C\to D$ preserving the colimits that $\mathrm{sk}$ is built from. If they are preserved (and I take it you assume that the dual limits that $\mathrm{cosk}$ is built from exist in $D$), then the desired relation will hold.

4. • CommentRowNumber4.
   • CommentAuthorMirco Richter
   • CommentTimeApr 19th 2012
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexNo the relative adjoint functor definition doesn't apply here. But ok, In that case I think the answer is just that you can't generally say that the natural isomorphism exists...

   No the relative adjoint functor definition doesn’t apply here. But ok, In that case I think the answer is just that you can’t generally say that the natural isomorphism exists…

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeApr 19th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItex> No the relative adjoint functor definition doesn't apply here. Here is how: Apply the following dictionary to translate from the notation at _[[relative adjoint functor]]_ to your notation * $C$ becomes $sD$ * $D$ becomes $sD$ * $B$ becomes $sC$ * $J : B \to D$ becomes $sC \hookrightarrow sD$. * $R : C \to D$ becomes $cosk_n : sD \to sD$ * $L : B \to C$ becomes $sk_n : sC \to sD$. With that the natural iso $$ Hom_C(L(-), -) \simeq Hom_D(J(-), R(-)) $$ becomes the natural iso that you are after $$ Hom_{sD}(sk_n(X), (Y)) \simeq Hom_{sD}(X, cosk_n Y) $$ for $X \in sC$, $Y \in sD$

   No the relative adjoint functor definition doesn’t apply here.

   Here is how:

   Apply the following dictionary to translate from the notation at relative adjoint functor to your notation

   • $C$ becomes $\mathrm{sD}$

   • $D$ becomes $\mathrm{sD}$

   • $B$ becomes $\mathrm{sC}$

   • $J:B\to D$ becomes $\mathrm{sC}\hookrightarrow \mathrm{sD}$.

   • $R:C\to D$ becomes ${\mathrm{cosk}}_n:\mathrm{sD}\to \mathrm{sD}$

   • $L:B\to C$ becomes ${\mathrm{sk}}_n:\mathrm{sC}\to \mathrm{sD}$.

   With that the natural iso

   $${\mathrm{Hom}}_C(L(-),-)\simeq {\mathrm{Hom}}_D(J(-),R(-))$$

   becomes the natural iso that you are after

   $${\mathrm{Hom}}_{\mathrm{sD}}({\mathrm{sk}}_n(X),(Y))\simeq {\mathrm{Hom}}_{\mathrm{sD}}(X,{\mathrm{cosk}}_{n}Y)$$

   for $X\in \mathrm{sC}$, $Y\in \mathrm{sD}$

6. • CommentRowNumber6.
   • CommentAuthorMirco Richter
   • CommentTimeApr 19th 2012
   • (edited Apr 19th 2012)
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexI know, but what I mean is that this is not a solution to the problem. Just a formal clean way to write it down. We only now that the natural iso exists in the subcategory $\mathbf{sC}$, but if I understand it right this doesn't implay $Hom_{sD}(sk_n(X),Y) \simeq Hom_{sD}(X,cosk_n Y)$ for $X \in sC$ and $Y \in sD$. Right? So I still ave to show that the iso exist here.

   I know, but what I mean is that this is not a solution to the problem. Just a formal clean way to write it down. We only now that the natural iso exists in the subcategory $\mathrm{sC}$, but if I understand it right this doesn’t implay

   ${\mathrm{Hom}}_{\mathrm{sD}}({\mathrm{sk}}_n(X),Y)\simeq {\mathrm{Hom}}_{\mathrm{sD}}(X,{\mathrm{cosk}}_{n}Y)$ for

   $X\in \mathrm{sC}$ and $Y\in \mathrm{sD}$. Right?

   So I still ave to show that the iso exist here.

7. • CommentRowNumber7.
   • CommentAuthorMirco Richter
   • CommentTimeApr 19th 2012
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexIf someone knows more please let me know. (I know most of the time I'm here for just asking questions, but you are the leading experts :-) and those things are quite cumbersome to me). The question is two fold: Assuming I know that in $sC$ the isomorphism $Hom_{sC}(sk_n(X),Y) \simeq Hom_{sC}(X,cosk_nY)$ exists. 1.) Do I still have to prove that there is a natural iso giving $Hom_{sD}(sk_n(X),Y) \simeq Hom_{sD}(X,cosk_n Y)$ for $X \in sC$ and $Y \in sD$, or is this automatic? 2.) If I still have to prove it, can I take somehow advantage of the fact that the isomorphism exist in the subcat $sC$?

   If someone knows more please let me know. (I know most of the time I’m here for just asking questions, but you are the leading experts :-) and those things are quite cumbersome to me). The question is two fold:

   Assuming I know that in $\mathrm{sC}$ the isomorphism ${\mathrm{Hom}}_{\mathrm{sC}}({\mathrm{sk}}_n(X),Y)\simeq {\mathrm{Hom}}_{\mathrm{sC}}(X,{\mathrm{cosk}}_{\mathrm{nY}})$ exists.

   1.) Do I still have to prove that there is a natural iso giving ${\mathrm{Hom}}_{\mathrm{sD}}({\mathrm{sk}}_n(X),Y)\simeq {\mathrm{Hom}}_{\mathrm{sD}}(X,{\mathrm{cosk}}_{n}Y)$ for $X\in \mathrm{sC}$ and $Y\in \mathrm{sD}$, or is this automatic?

   2.) If I still have to prove it, can I take somehow advantage of the fact that the isomorphism exist in the subcat $\mathrm{sC}$?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeApr 19th 2012
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   Author: Urs
   Format: MarkdownItexAs I said, you need that the inclusion preserves the colimits.

   As I said, you need that the inclusion preserves the colimits.

9. • CommentRowNumber9.
   • CommentAuthorMirco Richter
   • CommentTimeApr 20th 2012
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexOk. Just a little fine tuning: 1.) Must $J$ preserve all colimits or only those used in $sk_n$? 2.) Must $D$ have all limits or only those used in $cosk_n$?

   Ok. Just a little fine tuning:

   1.) Must $J$ preserve all colimits or only those used in ${\mathrm{sk}}_n$?

   2.) Must $D$ have all limits or only those used in ${\mathrm{cosk}}_n$?