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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?

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    $\begingroup$ 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
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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.)

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