Representations of a group and its normal subgroup While I was thinking the problem that I asked previously, I encountered this exercise problem in the book Tensor categories.
The problems (Exercise 4.15.3) is that

Let $N$ be a normal subgroup of a group $G$. Show that the quotient group $G/N$ acts on $\mathrm{Rep}(N)$ and $\mathrm{Rep}(N)^{G/N}\cong\mathrm{Rep}(G)$.

Here $\mathrm{Rep}(N)^{G/N}$ is the equivariantization of the category $\mathrm{Rep}(N)$ by the group $G/N$.
I have not understood what the equivariantization is very well. I understood that when $N$ is the trivial subgroup $\{e\}$ of $G$, then $\mathrm{Rep}(\{e\})^{G}\cong \mathrm{Rep}(G)$. But the general case, I couldn't prove it.
Is it related to the induction of a representation?
 A: I tried my best to be as clear as possible, but as often in category theory even the simple things are hard to write, so I hope this is readable (really there is nothing difficult in there).
Definition of the action
For any $g\in G$, write $\tilde{T_g}\in \mathrm{Aut}_\otimes(\mathrm{Rep}(N))$ defined by
$\tilde{T_g}(V,\rho) = (V,\rho_g)$ with $\rho_g: N\to \mathrm{Aut}_k(V)$ defined as $\rho_g(n) = \rho(g^{-1}ng)$ ; and $\tilde{T_g}$ is the identity on morphisms.
You can check that $\tilde{T_g}$ is indeed a tensor endofunctor of $\mathrm{Rep}(N)$ and that you have a strict equality $\tilde{T_{gh}} = \tilde{T_g}\circ \tilde{T_h}$ for all $g,h\in G$.
Then it's easy to see that if $g\in N$ then $\tilde{T_g}$ is isomorphic to the identity functor : just define $\alpha_g : \tilde{T_g}\to Id$ by $\alpha_g(V,\rho) = \rho(g) : V\to V$. This is because if $g\in N$ then we conjugate by an element of the group itself, so of course it's like conjugating by $\rho(g)$.
Now choose a set section $s: G/N\to N$ with $s(1)=1$, and let $c:(G/N)^2\to N$ be the usual "$2$-cocycle" (not a real cocycle if $N$ is not abelian, though) defined by $c(x,y) = s(xy)^{-1}s(x)s(y)$.
Then you can put, for any $x\in G/N$, $T_x = \tilde{T}_{s(x)}$. This way you get $T_x\circ T_y = T_{xy} \circ \tilde{T}_{c(x,y)}$. But in general if $\alpha: F\to F'$ in a functor isomorphism, then $G(\alpha):G\circ F\to G\circ F'$ is also a functor isomorphism. So here, with $G=T_{xy}$, $F = \tilde{T}_{c(x,y)}$ and $F'=Id$, you get an isomorphism of functors $\gamma_{x,y}: T_x\circ T_y\to T_{xy}$ defined by $$\gamma_{x,y}(V,\rho) = T_{xy}(\alpha_{c(x,y)}(V,\rho)) : (T_x\circ T_y)(V,\rho) = T_{xy}(\tilde{T}_{c(x,y)}(V,\rho)) \to T_{xy}(V,\rho)$$
(it's really painful to write but trivial to check). Note that as a linear map, $\gamma_{x,y}(V,\rho):V\to V$ is just $\rho(c(x,y))$.
Thus you get a well-defined tensor action of $G/N$ on $\mathrm{Rep}(N)$.
Equivariantization
Now we check that $\mathrm{Rep}(N)^{G/N}\simeq \mathrm{Rep}(G)$. 
What is an object of $\mathrm{Rep}(N)^{G/N}$ ? If you unravel the definition (you should probably do it to see how it works), given our action, it is an representation $(V,\rho)$ of $N$ with linear automorphisms $u_x$ of $V$ for all $x\in G/N$ satisfying, for all $n\in N$, $\rho(s(x)^{-1}ns(x)) = u_x^{-1}\circ \rho(n)\circ u_x$ and $u_x\circ u_y = u_{xy}\circ \rho(c(x,y))$ (the first equality means that $u_x$ is a morphism of representations from $T_x(V,\rho)$ to $(V,\rho)$, and the second is the commutative square relating the $u_x$ with $\gamma_{x,y}$ given in your book).
If $(V,\rho)$ is a representation of $G$, then the obvious choice is $u_x = \rho(s(x))$.
Formally, write $R: \mathrm{Rep}(G)\to \mathrm{Rep}(N)$ for the usual restriction functor. There is an "obvious" functor $F: \mathrm{Rep}(G)\to \mathrm{Rep}(N)^{G/N}$ such that $F(X) = (R(X), u)$ for all $X=(V,\rho)$ in $\mathrm{Rep}(G)$, with for all $x\in G/N$, $$u_x = R(\alpha_{s(x)}(V,\rho)) : T_x(R(V,\rho)) = R(\tilde{T}_{s(x)}(V,\rho))\to R(V,\rho).$$
A little comment on that formula : when we say $\alpha_{s(x)}(V,\rho)$ and $\tilde{T}_{s(x)}(V,\rho)$, it's because $(V,\rho)$ is a representation of $G$ and $s(x)\in G$ so based on what is above $\alpha_{s(x)}$ is a functor isomorphism between $\tilde{T}_{s(x)}$ and $Id$, which are endofunctors of $\mathrm{Rep}(G)$ (again, tiresome but ultimately trivial). 
In the other direction, suppose we have $((V,\rho),u)$ with $(V,\rho)$ representation of $N$, and $u_x: T_x(V,\rho)\to (V,\rho)$ isomorphisms that satisfy the coherence condition.
We want a morphism $\varphi: G\to \mathrm{Aut}_k(V)$ that extends $\rho$. We can already define $\varphi(g) = \rho(g)$ if $g\in N$.
For any $x\in G/N$, we define $\varphi(s(x)) = u_x : V\to V$ which is quite natural given that $\rho(s(x)^{-1}ns(x)) = u_x^{-1}\circ \rho(n)\circ u_x$.
Then for any $g\in G$, we write $g = s(\overline{g})n(g)$ where $\overline{g}\in G/N$ is the class of $g$ and $n(g)\in N$, and we define $\varphi(g) = \varphi(s(\overline{g}))\circ \varphi(n(g))$.
We must check that this $\varphi$ is a well-defined group morphism (for now it's a well-defined function). Let $g,h\in G$. We have by definition $\varphi(gh) = u_{\overline{gh}}\circ \rho(n(gh))$ and $\varphi(g)\varphi(h) = u_\overline{g}\circ \rho(n(g)) \circ u_\overline{h}\circ \rho(n(h))$. Now you can directly check that $n(gh) = c(\overline{g},\overline{h})\cdot (s(\overline{h})^{-1}n(g)s(\overline{h}))\cdot n(h)$, so 
$$\begin{eqnarray*} 
u_{\overline{gh}}\circ \rho(n(gh)) & = & u_{\overline{gh}}\circ \rho(c(\overline{g},\overline{h})) \circ \rho(s(\overline{h})^{-1}n(g)s(\overline{h}))\circ \rho(n(h)) \\ 
& = & u_\overline{g}\circ u_\overline{h}\circ (u_\overline{h}^{-1}\circ \rho(n(g))\circ u_\overline{h})\circ \rho(n(h)) \\
 & = & \varphi(g)\varphi(h). \end{eqnarray*}$$
I leave it to you to check that $F$ and $G$ are indeed functors (I did not define them on morphisms but it should be easy) and that $G\circ F\simeq Id$ and $G\circ F\simeq Id$, giving the equivalence of categories.
