Consider the theorem at the end of chapter six:

Theorem: Let $\mathbf{I}$ be a small category and $\mathbf{A}$ a category with all limits of shape $\mathbf{I}$. Then $\lim_{\mathbf{I}}$ defines a functor $[\mathbf{I}, \mathbf{A}] \rightarrow \mathbf{A}$ which is right-adjoint to the diagonal functor $\Delta : \mathbf{A} \rightarrow [\mathbf{I}, \mathbf{A}]$.

We wish interpret this theorem in the special case that $\mathbf{A} = \mathbf{Set}$ and $\mathbf{I}$ is a group.

First note that $[G, \mathbf{Set}]$ is the category of $G$-sets -- that is, groups equipped with a $G$-action $G \times S \rightarrow S$. In this scenario, the diagonal functor $\Delta : \mathbf{Set} \rightarrow [G, \mathbf{Set}]$ takes a set $S$ to $(S, \cdot)$, where $S$ is now equipped with the trivial $G$-action where $g \cdot s = s$ for all $g$ and $s$ (alternatively, $g \cdot \dash = \id_S$ for all $g$).

To find out what $\lim_{\mathbf{I}}$ does, we first need to consider what constitutes a cone over a $G$-set. Given a $G$-set $(S, \cdot)$, a cone consists of a set $A$ and a single map $f : A \rightarrow S$ such that $(g \cdot \dash) \circ f = f$ for all $g$. To rephrase, that means that $g \cdot f(a) = f(a)$ for all $a \in A$ and $g \in G$. This means that every $f(a)$ is stabilized by the whole group $G$. So a cone is a pair $(A, f)$ where $f$ must take each element to a $G$-invariant element. So a universal cone (limit cone) can be computed by the usual construction giving arbitrary limits in $\mathbf{Set}$. We find that the limit is precisely the subset of $S$ consisting of $G$-invariant elements.

We are finally ready to interpret the adjunction. The adjunction means that $[G, \mathbf{Set}](\Delta S, T) \cong \mathbf{Set}(S, \lim_{\mathbf{I}} T)$ for any set $S$ and $G$-set $(T, \cdot)$. Indeed, one can observe that a $G$-equivariant map from $\Delta S$ to $T$ can only map onto $G$-invariant elements of $T$, so of course such a map is determined completely by a map $S \rightarrow \lim_{\mathbf{I}} T$, which is precisely the subset of $G$-invariant elements.

The dual of the preceding theorem says that colimits are left-adjoint to the diagonal functor. In the present context, let's interpet the $\colim_{\mathbf{I}}$ functor. Consider some $G$-set $(S, \cdot)$. We can use the general construction for colimits in set and find that the colimit is $S / {\sim}$, where $\sim$ is the precisely the equivalence relation induced by the $G$-action (orbits correspond to equivalence classes). So, the adjunction says that $\mathbf{Set}(S / {\sim}, T) \cong [G, \mathbf{Set}](S, \Delta T)$. Indeed, its easy to see that for a $G$-equivariant map $S \rightarrow \Delta T$, all members of an orbit must map on to the same element, so of course such a map is determined by a mapping from the orbits of $S$ to $T$.