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I am reading a paper "2-vector spaces and groupoid" by Jeffrey Morton and I need a help to understand the following.

Let $X$ and $Y$ be finite groupoids. Let $[X, \mathbb{Vect}]$ be a functor category from $X$ to $\mathbb{Vect}$. Suppose we have a functor $f: X \rightarrow Y$. We want to construct a pushforward $f_{*} :[X, \mathbb{Vect}] \rightarrow [Y, \mathbb{Vect}]$.

For each object $y \in Y$, the comma category $(f \downarrow y)$ has object which are objects $x \in X$ with maps $f(x) \rightarrow y$ in $Y$, and morphisms which are morphisms $a: x \rightarrow x'$ whose images make the evident triangle in $Y$ commute.

Then the author defines for each $F\in [X, \mathbb{Vect}]$, $f_{*}(F)(y):=colim F(f \downarrow y)$.

He also shows that $f_{*}(F)(y)=\bigoplus_{f(x)\cong y} \mathbb{C}[Aut(y)]\otimes_{\mathbb{C}[Aut(x)]}F(x)$.

The first thing I don't understand is the definition of $ colim F(f \downarrow y)$. Is it a colimit of a functor $F: (f \downarrow y) \rightarrow \mathbb{Vect}$ by regarding the comma category as a subcategory of $X$?

The second thing I don't understand is the proof of the explicit formula above. The author first consider the case when $X$ and $Y$ are just finite groups $G$ and $H$ respectively and $f :G \rightarrow H$ is a homomorphism and we have a representation of $G$ on a vector space $V$. In this case we have $f_{*}V=\mathbb{C}[H] \otimes_{\mathbb{C}[G]} V$.

I don't know how to calculate the colimit above and deduce this formula.

I appreciate any help. Thank you in advance.

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    The author has abused notation: what he really means is to take the colimit of the functor obtained by composing the obvious projection $(f \downarrow y) \to X$ with the given presheaf $F : X \to \textbf{Vect}$. This is also known as the left Kan extension of $F : X \to \textbf{Vect}$ along the functor $f : X \to Y$.2012-10-22

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The answer to your first question is roughly yes, although in general a slice category may not be a subcategory since the image of an object in the source can admit many maps to the target, and these all give distinct (possibly non-isomorphic) objects in the slice. Of course, you can still define the pullback along the forgetful functor.

As for your second question, this is just the fact that the "tensoring up" functor $\mathbb{C}[H] \otimes_{\mathbb{C}[G]} - :\mbox{Rep}(G) \rightarrow \mbox{Rep}(H)$ is adjoint to the forgetful (or "restriction of structure") functor $U:\mbox{Rep}(H) \rightarrow \mbox{Rep}(G)$.

To actually compute this colimit, let's write $\mathcal{G}$ and $\mathcal{H}$ for the categories, $*_G$ and $*_H$ for their unique objects, and identify the morphisms with the sets $G$ and $H$. Then the objects of $(f\downarrow *_H)$ are given by pairs $(*_G,h:*_H \rightarrow *_H)$ (for all $h\in H$), and the morphisms of $(f\downarrow *_H)$ are given by $$ \mbox{Hom}_{(f\downarrow *_H)}((*_G,h_1:*_H \rightarrow *_H),(*_G,h_2:*_H \rightarrow *_H) = \{g\in G : f(g)=h_2^{-1}h_1\} $$ (using the standard convention for composition).

Now we can explicitly describe the diagram $(f \downarrow *_H) \xrightarrow{U} \mathcal{G} \xrightarrow{F} \mbox{Vect}$: it takes every object to $V$, and to the morphism $f(g)=h_2^{-1}h_1$ it associates the linear homomorphism (actually isomorphism) $F(g)$ from the copy of $V$ over $h_1$ to the copy of $V$ over $h_2$.

Colimits are built as follows: given a diagram $D:\mathcal{I} \rightarrow \mathcal{C}$, $$ \mbox{colim}(D) \cong \mbox{coeq} \left( \coprod_{(i_0 \rightarrow i_1)\in \mbox{mor}(\mathcal{I})} D(i_0) \rightrightarrows \coprod_{j \in \mbox{ob}(\mathcal{I})} D(j) \right),$$ where one map takes the $D(i_0)$ over $(i_0 \rightarrow i_1)$ to $D(i_1)$ over $j=i_1$ via $D(i_0 \rightarrow i_1)$, and the other map takes $D(i_0)$ over $(i_0 \rightarrow i_1)$ to $D(i_0)$ over $j=i_0$ via the identity map. Of course, in $\mbox{Vect}$ all this means is that you take a big direct sum of all the vector spaces in the diagram and then take the quotient where we identify $v\in D(i_0)$ with $(D(i_0 \rightarrow i_1))(v) \in D(i_1)$ (for all $(i_0 \rightarrow i_1) \in \mbox{mor}(\mathcal{I})$).

So for our situation, we start with the direct sum of a bunch of copies of $V$ indexed by $H$, i.e. $\bigoplus_H V \cong \mathbb{C}[H] \otimes V$. (Categorically, one would say that $\mbox{Vect}$ is tensored (or copowered) over $\mbox{Set}$, which essentially means that we have a reasonable and (bi)functorial notion of "taking coproducts of an object indexed by a set".) Then, if we have a vector $v\in V$ considered as sitting over $h_1$ -- that is, the element $h_1 \otimes v$ -- and $f(g)=h_2^{-1}h_1$ is a morphism from $h_1$ to $h_2$, then we make the identification $h_1 \otimes v \sim h_2 \otimes g \cdot v$. But notice that this is just the usual definition of trading off the right action of $\mathbb{C}[G]$ on $\mathbb{C}[H]$ via $f$. (To be precise, this action comes from the functoriality and the product-preservation of the tensored structure I mentioned above.) So finally, $\mbox{colim}_{(f \downarrow *_H)} F \circ U \cong \mathbb{C}[H] \otimes_{\mathbb{C}[G]} V$.

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    Could you tell me how to get the formula from the colimit definition?2012-10-22
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    There you go. It's super late here and so it's easily possible that I'm being needlessly complicated and/or pedantic and/or categorical, but hopefully you find it helpful.2012-10-22
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    Colimits in the category of vector spaces are _not_ built using tensor products. As usual they are built using coproducts and coequalisers.2012-10-22
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    Ugh, I figured doing this in the middle of the night was a bad idea. Thanks, I'll edit.2012-10-22
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    Thank you very much. I really appreciate your answer. I'm getting understand it but it seems that I am too weak at category theory. I didn't know that colimits can be calculated by coproducts and coequalizers. Could you suggest me a text book to study these category theory?2012-10-22
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    Yes, I've used the general definition of colimit of a diagram above; for smaller diagrams (e.g. a pushout) one often has an easy and explicit formula, but when you don't actually know the shape of your diagram then you need to be more general. I've never read it, but everyone recommends MacLane's "Categories for the Working Mathematician", so I guess I'd point you there.2012-10-22
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    Of course, let me know if I can clarify anything in my answer above.2012-10-22
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    @AaronMazel-Gee Thank you very much. I like your explanation. One thing I want to ask is about the $\bigoplus_H V \cong \mathbb{C}[H] \otimes V$. Why do we use a group algebra here? Can we say that $\bigoplus_H V \cong H \otimes V$ instead?2012-10-22
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    Haha. Well so this "tensoring" business has to do with the bifunctor $\mbox{Set} \times \mbox{Vect} \rightarrow \mbox{Vect}$ given by $(S,V) \mapsto \mathbb{C}[S] \otimes V$. In general, $\mathbb{C}[S]$ is *not* a group algebra, but just a vector space with a distinguished basis. In fancy categorical language, you *could* write $S \otimes V$ -- this is standard notation for any tensored structure $\mathcal{V} \times \mathcal{C} \rightarrow \mathcal{C}$ -- but here I think that would mostly just be confusing.2012-10-22
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    To say that $\mathbb{C}[H]$ is a $\mathbb{C}$-algebra is to say that it's a *monoid object* (in fact, a *group object*) in the category $\mbox{Vect}$. This comes from the facts that (a) $H$ is a monoid object in $\mbox{Set}$, and (b) the restriction $\mathbb{C}[-]$ of the above bifunctor to $\mathbb{C}$ in the second slot is product-preserving.2012-10-22
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    Let me write $\boxtimes$ for the categorical tensor structure for a moment (so that $S \boxtimes V = \mathbb{C}[S] \otimes V$). Then, the categorical reason we choose $\mathbb{C}$ is that it is the *unit* for the monoidal category $(\mbox{Vect},\otimes)$; that is, there are natural isomorphisms $\mathbb{C} \otimes V \cong V$. So, this unit object allows us to "bring objects of $\mbox{Set}$ into $\mbox{Vect}$ for free", in the sense that $(S \boxtimes -) \cong ((S \boxtimes \mathbb{C}) \otimes -):\mbox{Vect} \rightarrow \mbox{Vect}$.2012-10-22
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    (A category $\mathcal{C}$ that's tensored over $\mathcal{V}$ might also admit some tensor-product-like bifunctor $-\otimes - : \mathcal{C}\times \mathcal{C} \rightarrow \mathcal{C}$, such that we have natural isomorphisms $(V \boxtimes X) \otimes Y \cong V \boxtimes (X \otimes Y)$. But if the functor $-\otimes -$ doesn't have a unit, then you won't be able to do the above "bring objects of $\mathcal{V}$ into $\mathcal{C}$ for free" construction. (Note that the existence of a unit object is generally taken to be part of the definition of a monoidal category.))2012-10-22
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    @AaronMazel-Gee Thank you. I am wondering how you have studied these category theory.2012-10-24
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    @Primo: A lot of it I've just picked up in the course of reading. Conversations are a big part of my learning process, too. The nLab wiki is also an excellent resource for category theory and all sorts of related things, although they generally tend to phrase everything with an eye towards $\infty$-categories, so it can be a bit dense and overly formal. I guess I'd recommend continuing to focus on other things that interest you, and picking up category theory as necessary in along the way; that technique has been working for me, at least.2012-10-24
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    Certainly others might disagree with this approach. But personally, I find category theory natural enough that when I learn something about it, it generally tends to stick. At least for the basics, there's never anything to do in category theory proofs besides follow your nose.2012-10-24