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Let $k$ be a field and $G$ a group. Consider $G-\mathrm{vect}_k$ the category of $G$-graded vector spaces (compare nLab). I have the following proposition to prove:

For a field $k$ and a finite group $G$ the following holds for the categories $\mathrm{vect}_k$ and $G-\mathrm{vect}_k$: $$(G-\mathrm{vect}_k)_0 \cong \mathrm{vect}_k$$

I suppose the equivalence is meant to be as monoidal categories, as both have the structure of a monoidal category with the normal vector space tensor product. I understand the basic idea behind this, as vector spaces in $\mathrm{vect}$ have no grading and in every component every vector space can be considered. But I have problems with formalizing this. Can anyone maybe help me to find a start formulating this?

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  • $\begingroup$ What is your definition of $(G-\mathrm{vect}_k)_0$ ? It doesn't seem to appear on the nLab page you linked to $\endgroup$ – Max Jul 8 at 11:07
  • $\begingroup$ I expected it to be all the elements of grade zero as it is used here for one vector spaces: en.wikipedia.org/wiki/Graded_vector_space. $\endgroup$ – P. Schulze Jul 8 at 11:17
  • $\begingroup$ Oh so like the objects of $G-\mathrm{vect}_k$ that are concentrated in degree $0$ ? $\endgroup$ – Max Jul 8 at 11:19
  • $\begingroup$ what do you mean by concentrated? $\endgroup$ – P. Schulze Jul 8 at 11:19
  • $\begingroup$ that $V_x=0$ unless $x=e$ ($e$ neutral element) $\endgroup$ – Max Jul 8 at 11:22
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If I'm correct in my interpretation that $(G-\mathrm{vect}_k)_0$ is the full subcategory of $G-\mathrm{vect}_k$ on objects that are concentrated in degree $e$, that is objects $V$ such that $\forall x, (x=e \lor V_x=0)$, then the claim is relatively easy.

Indeed, if you have two of these, $V,W$ then a map $V\to W$ is a collection of maps $f_x: V_x\to W_x$. But for all $x$ except $e$, $V_x=W_x= 0 $ so $f_x=0$: the only interesting one is $f_e$. Similarly, since we know all the other $V_x$'s (they're $0$ !) we only care about $V_e$. This shows intuitively why essentially this is the same as $\mathrm{vect}_k$.

Formally, you can define a functor $(G-\mathrm{vect}_k)_0\to \mathrm{vect}_k$ via $V\mapsto V_e$, $f:V\to W\mapsto f_e :V_e\to W_e$, check that it is fully faithful and essentially surjective (in fact, depending on how you define things, it's even literally surjective and injective, but this depends on the specific details of implementation); so it is an equivalence (you can even define a quasi-inverse quite easily !)

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  • $\begingroup$ just to be sure: The quasi inverse could be the functor $V \mapsto V=\oplus V_g$ with $V_g = 0 \forall g \in G\{0}$ and $V_e = V$ ? $\endgroup$ – P. Schulze Jul 8 at 14:13
  • $\begingroup$ On objects, yes $\endgroup$ – Max Jul 8 at 14:17

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