How can I see that the slice of a presheaf category is equivalent to the presheaf category of the category of elements? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:11:58Z http://mathoverflow.net/feeds/question/89635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89635/how-can-i-see-that-the-slice-of-a-presheaf-category-is-equivalent-to-the-presheaf How can I see that the slice of a presheaf category is equivalent to the presheaf category of the category of elements? Chuck 2012-02-27T05:10:05Z 2012-02-27T10:39:00Z <p>Let $\mathcal{C}$ be a small category and $P$ a presheaf on $\mathcal{C}$. Then there is an equivalence $$\widehat{\mathcal{C}} / P \cong \widehat{\int_{\mathcal{C}}} P$$ Now there are (at least) two ways to see this (from what I've managed to ascertain.) You might derive the equivalence as it were 2-categorically using the universal property of $\widehat{\mathcal{C}}$ which is the equivalence $\textbf{Cat}/\widehat{\mathcal{C}} = \textbf{Func }(\widehat{\mathcal{C}}) \cong \textbf{DFib}(\mathcal{C})$ where $A$ is any small category and $\textbf{DFib}$ are the discrete fibrations over $\mathcal{C}$. You do this by showing that any functor $A \rightarrow \widehat{\mathcal{C}}/P$ corresponds to a functor $A \rightarrow \widehat{\mathcal{C}}$ and a natural transformation $A \rightarrow 1 \overset{P}{\rightarrow} \mathcal{C}$ (where $1$ is terminal in $\textbf{Cat}$) and hence that the corresponding morphism of fibrations corresponds to a discrete fibration of ${\int_{\mathcal{C}}}P$ and hence that </p> <p>$$\textbf{Func }(\widehat{\mathcal{C}}/P) \cong \textbf{DFib}(\int_{\mathcal{C}}P) \cong \textbf{Func }(\widehat{\int_{\mathcal{C}}P})$$ </p> <p>which by universality gives the desired equivalence. Now I have to admit I'm a bit sketchy on the details of this, as it uses many non-obvious properties of fibrations. A more detailed version of this approach was given as a partial answer to <a href="http://mathoverflow.net/questions/23887/slices-of-presheaf-categories" rel="nofollow">this question</a>.</p> <p>I am more interested in the more elementary approach, i.e. establishing an explicit equivalence between the two categories, so that I can understand, morally, why this equivalence ought to hold. The only natural functor I've been able to find between the two is given firstly by factorizing $$\int_{\mathcal{C}}P \overset{\pi_P}{\rightarrow} \mathcal{C} \overset{y}{\rightarrow} \widehat{\mathcal{C}}$$ through the Yoneda embedding to $\widehat{\int_{\mathcal{C}}P}$ and the canonical colimit functor $L : \widehat{\int_{\mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}$ given since $y \circ \pi_P$ is a functor to a cocomplete category, and then getting to $\widehat{\mathcal{C}}/P$ by pullback $p$ along $P \rightarrow 1$. So the end result is a functor $$p \circ L: \widehat{\int_{\mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}/P$$ Does this functor provide the desired equivalence? There seems to be significant loss of information at $p$, but on the other hand $L$ does have the property that it makes the two respective Yonneda embeddings commute (i.e. $L \circ y = y \circ \pi_P$), so morally it seems that the information lost at $p$ should be the same amount of information lost by the projection $\pi_P : \int_{\mathcal{C}}P \rightarrow \mathcal{C}$ (which ought to make $p \circ L$ full and faithful; and I think essential surjectivity is straightforward.)</p> <p>But I also get the sense I am chatting nonsense. If this functor does nothing of the sort then the question, I suppose, is what is the most elementary functor that provides this equivalence? </p> <p>Any helpful comments will be greatly appreciated both by myself and my bloodshot eyes.</p> http://mathoverflow.net/questions/89635/how-can-i-see-that-the-slice-of-a-presheaf-category-is-equivalent-to-the-presheaf/89648#89648 Answer by Martin Brandenburg for How can I see that the slice of a presheaf category is equivalent to the presheaf category of the category of elements? Martin Brandenburg 2012-02-27T10:26:52Z 2012-02-27T10:39:00Z <p>You can verify this equivalence elementarily (without the language of fibrations etc.):</p> <p>Assume first that $C$ is the category with only one morphism (i.e. the terminal category), so that a presheaf on it is just a set. Then the statement is as follows: If $P$ is a set, then a set $F$ together with a map $F \to P$ is the same as to give a family of sets indexed by $P$. But this is obvious, right? Given $F \to P$, we may look at its <em>fibers</em> $F_s$, where $s$ runs through the elements $s \in P$. Since $F$ is the disjoint union of the $F_s$, it is also clear how to write down the inverse.</p> <p>For general $C$ it works in the same way, $C$ is just a sort of parametrization.</p> <p>If $F \in \widehat{C} / P$, i.e. $F$ is a presheaf on $C$ together with a morphism $F \to P$, then define a presheaf $G$ on $\int_C P$ as follows: If $(X,s)$ is an element of $P$, i.e $s \in P(X)$, then let $G(X,s)$ be the fiber of $s$ with respect to $F(X) \to P(X)$.</p> <p>Conversely, if $G$ is a presheaf on $\int_C P$, then define a presheaf $F$ on $C$ as follows: For $X \in C$ let $F(X) = \coprod_{s \in P(X)} G(X,s)$. We have a natural projection $F(X) \to P(X)$, which gives rise to a morphism $F \to P$.</p> <p>Now it is straight forward to check that these assignments actually define functors which are pseudo-inverse to each other.</p>