Let $(G,\cdot)$ be a group and let $BG$ be the category of this group with one formal object $*$ and the elements of $G$ as morphisms.

Now take the the covariant hom-functor $\text{Hom$(*,\_)$}:BG \to \mathbf{Set}$

$*\mapsto \text{Hom$(*,*)$} = [\text{the set of morphisms from $*$ to $*$ }] = \{g\in G\} = G$, seen as a set

Now to my question and confusion:

Let $g\in G$ then,

$$g \mapsto \text{Hom$(*,g)$} $$

This is $$\text{Hom$(*,g)$}:\text{Hom$(*,*)$} \to \text{Hom$(*,*)$}$$ $$\_ \mapsto g\circ \_$$

If $\text{Hom$(*,*)$}$ is the group seen as a set then what is $\text{Hom$(*,g)$}$? Is it endomorphisms or even bijections of $G$? with left multiplication of $g$? Can anyone please explain how this works?

  • 1
    $\begingroup$ A common notation for the category associated to $G$ is $BG$. (The nerve of this category is the usual Bar resolution of the classifying space.) $\endgroup$ – Martin Brandenburg Nov 3 '14 at 17:02
  • $\begingroup$ @MartinBrandenburg Thank you! I wasn't aware of that. $\endgroup$ – John Smith Nov 3 '14 at 17:02

In general, if $X$ is an object and $g : Y \to Z$ is a morphism, then $\hom(X,Y) \to \hom(X,Z)$ is defined by $h \mapsto g h$. This doesn't change if $X$ is the only object. Hence, your map is left multiplication with $g$ (this is a bijection, but not a homomorphism unless $g=1$). Hence, the functor $\hom(\star,-)$ is the left regular representation of $G$.

(Now it is a good exercise to deduce Cayley's Theorem from the Yoneda Lemma.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.