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The Proposition appearing in the wonderful answer of Zhen Lin to this other question states that for a small category $\mathbb{C}$ and an object $X\in \mathbb{C}$

(*)\begin{equation} \widehat{\mathbb{C}}\downarrow Y(X) \cong \widehat{\mathbb{C}\downarrow X} \end{equation}

i.e. the category of presheaves on $\mathbb{C}$ over the Yoneda embedded $X$ is equivalent (even isomorphic?) to the category of presheaves on $\mathbb{C}\downarrow X$. This appears also in Sheaves in Geometry and Logic by MacLane and Moerdijk, Exercise III.8.

There is an obvious functor $(-)|_X:\widehat{\mathbb{C}}\to \widehat{\mathbb{C}\downarrow X}$ by precomposing a presheaf $\mathbb{C}^{op}\to Set$ with the opposite of the forgetful functor $\mathbb{C}\downarrow X\to \mathbb{C}$.

  • Can this functor $(-)|_X$ be identified with the isomorphism $F$ from the left to the right in (*)? If not, how do $F$ and $(-)|_X$ relate?

If you start with an object $p:Z\to Y(X)$ of $\widehat{\mathbb{C}}\downarrow Y(X)$ and apply $(-)|_X$, you get $Z|_X\to Y(X)_X$ but $Y(X)_X=Hom_{\mathbb{C}\downarrow X}(-,X)$ is the terminal presheaf on $\mathbb{C}\downarrow X$. Hence it seems to me that the structure morphism $p$ is not respected by $(-)|_X$, we only get $Z|_X$.

On the other hand, the last page of chapter VII of the above mentioned book (or Exercise III.8.b) states the equivalence (*) for sheaves where $\mathbb{C}\downarrow X$ carries the obvious Grothendieck topology induced from $\mathbb{C}$. Doesn't this somehow suggest that the functor $(-)_X$ plays a role in the equivalence?

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1 Answer 1

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Good observation. The answer is that $(-)|_X$ does induce the equivalence of categories, but in a subtle way.

Let $\mathbb{D} = (\mathbb{C} \downarrow X)$ and let $u = (-)|_X$. First of all, there is an induced functor $u^* : \hat{\mathbb{C}} \to \hat{\mathbb{D}}$ defined by precomposing a presheaf on $\mathbb{C}$ with the functor $u$. Let $u_! : \hat{\mathbb{D}} \to \hat{\mathbb{C}}$ be the left Kan extension. Given a presheaf $F$ on $\mathbb{D}$, $u_! F$ is defined to be the colimit in $\hat{\mathbb{C}}$ of a canonical diagram of shape $\int^{\mathbb{D}} P$, and one may verify that in fact we have the explicit formula $$u_! F (C) = \{ (f, z) : f \in \mathbb{C}(C, X), z \in F (f) \}$$ In particular, if $F$ is the terminal object in $\hat{\mathbb{D}}$, $u_! F$ is isomorphic to $H_X = \mathbb{C}(-, X)$. Thus $u_!$ lifts to a functor $\hat{\mathbb{D}} \cong (\hat{\mathbb{D}} \downarrow 1) \to (\hat{\mathbb{C}} \downarrow H_X)$, and this is the required equivalence of categories.

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Wow, that's tricky. So there is a composition of two left adjoints $ u_{!}=forget\circ u'_{!}:\widehat{\mathbb{D}}\to (\widehat{\mathbb{C}}\downarrow X)\to \widehat{\mathbb{C}}$ and I guess that either $forget$ and $u'_{!}$ do have right adjoints. For $forget$, I would suppose $(-\times X)$ (I ingonre the Yoneda embedding in the notation) and for $u'_{!}$ it's the one you describe in Proposition to the cited answer, let's denote it by $v$. But does this mean that $u^{*}$ is the composition $v\circ (-\times X)$? All very confusing... –  geometrystudent Aug 8 '12 at 15:55
Yes, that's correct. It is a general fact that the left adjoint of the composite of two right adjoints is (up to natural isomorphism) the composite of the two left adjoints. –  Zhen Lin Aug 9 '12 at 2:04

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