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Recall the definition of the join of two simplicial sets. We may regard the functor $-\star Y$ as a functor $i_{Y,-\star Y}:Set_\Delta\to (Y\downarrow Set_\Delta)$ by replacing the resulting simplicial set $X\star Y$ with the canonical map $i_{Y,X\star Y}: Y\cong \emptyset\star Y \to X\star Y$. It is not hard to show that $i_{Y,-\star Y}$ commutes with colimits in $(Y\downarrow Set_\Delta)$, and therefore that it admits a right adjoint. We call the adjoint functor the overcategory or over-simplicial-set functor. We can give an explicit description of this simplicial set: Given an arrow $f:Y\to S$, define $(S\downarrow f)_n:=Hom_{(Y\downarrow Set_\Delta)}(i_{Y,\Delta^n\star Y},f)$.

As with all adjunctions, we have a unit and counit natural transformation $\eta_X:X\to (X\star Y\downarrow i_{Y,X\star Y})$, and $\epsilon_f: i_{Y,(X\downarrow f)\star Y}\to f$ respectively.

Then the question: Intuitively, the unit map should map $X$ to the simplicial subset of $(X\star Y \downarrow i_{Y,X\star Y})$ spanned by the "original" vertices of $X$, and I have seen this applied before (specifically in the case where $X$ is a simplex). However, I can't seem to figure out why this should be true formally. Is it true, and if so, how can we prove it?

Edit: As Tim mentions, the augmentation is given by the empty presheaf, that is, $\Delta^{-1}:=\emptyset$, which gives $X(-1):=Hom(\Delta^{-1}, X)=Hom(\emptyset,X)=\{\ast\}$

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Harry: I suppose there is a colon missing in the second line., but in addition, usually the join is defined for augmented simplicial sets. are you assuming some canonical augmentation and if so which one? –  Tim Porter Sep 7 '10 at 12:01
Why the vote down? –  Harry Gindi Sep 8 '10 at 1:58
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up vote 2 down vote accepted

It may help to look at Phil Ehlers thesis ``Algebraic Homotopy in simplicially Enriched groupoids'' Bangor 1993. He used some ideas from Duskin and van Osdol and write down the details of the right adjoint.(It can be found on the n-Lab at http://ncatlab.org/nlab/files/Ehlers-thesis.pdf )

I do not know if that will answer your question, since it is quite a long time since I read it, but he did describe the adjunction explicitly (see his chapter 3). The restriction placed in the answer to your earlier MO question (namely that $X(-1) = *$).

The right adjoint is given by $[Y,Z]_{n-1} = Set_{\Delta_a}(Y,Dec^nZ)$. The derivation is just jiggling about using the end calculus.

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