• One way to think about labellings is that we track the "entire history" of the object.
• it's hard to count unlabelled objects. it's easier to count labelled objects.
• for example, suppose we have graphs $g = (v, e)$ and $h = (v', e')$. an isomorphism of these as unlabelled graphs is a bijection function $f: v \rightarrow v'$ such that $s e t$ if and only if $f(s) e f(t)$.
• there could be many such $f$, or no such $f$. it's hard to find out!
• Now let's suppose the graphs have labellings, so we have labels $l: V \rightarrow [|V|]$ and $l': V' \rightarrow [|V'|]$where $[n] \equiv \{1, 2, \dots, n\}$.
• An isomorphism of labelled graphs is an unlabelled isomorphism along with the constraint that $l'(f(v)) = l(v)$. That is, we must preseve labels. So, for example, the graph:

a:1 -- b:2c:2 -- d:1

are isomorphic since I can send a -> d and b -> c.

• On the other hand, the graph:

a:1-b:2-c:3d:1-e:3-f:2

is not isomorphic (though they would be if we forget the numbering), since the center vertices b and e have different labels.

• Let's think of the equation $l'(f(v)) = l(v)$. Since $f$ is a bijection, we have $|V| = |V'|$, so $l$ and $l'$ are both bijections to the same set $[|V|] = [|V'|]$. So we can invert the equation to write $f(v) = l'^{-1}(l(v))$. This tells us that $f$ is determined by the labellings!
• The point of having a labelling is that it forces upon us a unique isomorphism (if it exists), given by the equation $f(v) \equiv l^{-1}(l(v))$.
• This collapses hom sets to either empty, or a unique isomorphism, which is far tamer than having many possible graph isomorphisms that we must search for / enumerate !
• In analogy to representation theory, if we consider two irreducible representations of a group $G$, say $\alpha: G \rightarrow GL(V)$ and $\beta: G \rightarrow GL(W)$, Schur's lemma tells us that the Hom-set between the two representations (an intertwining map) is either the zero map (which is like having no isos) or a scaling of the identity map (which is like having a uniquely determined iso).
• In this sense, we can think of an irrep as a "labelling" of group elements in a particularly nice way, since it constrains the potential isomorphisms of the "labelled objects"!