2
$\begingroup$

Consider a functor $F:A\to B$. It is an object in category $B^A$.

Question: is $B^A(F,F)$ a singleton?

To put is in other words: do we have non-identity natural transformations $\tau:F \to F $?

  • 0
    It's not quite standard notation. In algebraic topology one writes $\mathrm{B} G$ for the Eilenberg–MacLane space $K(G, 1)$, and there is a bijection between the set of $G$-torsors over a sufficiently nice space $X$ and the set of homotopy classes of continuous maps $X \to \mathrm{B} G$. In topos theory, there is a topos $\mathbf{B} G$ such that there is an equivalence of categories between geometric morphisms $\mathcal{E} \to \mathbf{B} G$ and internal $G$-torsors in $\mathcal{E}$. The topos $\mathbf{B} G$ turns out to be $[\mathbb{B} G^\textrm{op}, \textbf{Set}]$.2012-06-03

2 Answers 2

4

I hope Andy won't be bothered if I answer witht he "general case" of his construction. Take any category $\mathbf C$ and the hom-functor $F=\hom(X,-)$; then by Yoneda Lemma $ Nat(F,F)\cong \hom(X,X) $ Now you just have to choose "properly" your $X$ (e.g. by taking an object with a non-trivial endomorphism monoid).

  • 0
    Thank you, pleaae see my comment above2012-06-03
3

Consider $\mathrm{Set}^{\mathrm{M}^{\operatorname{op}}}$, where $M$ is a group viewed as a category. This is equivalent to the category of $\mathrm{Set-M}$ of right $M$ actions. If you write out the commutative diagram for a natural transformation for a functor $F\colon \mathrm{M}^{\operatorname{op}} \to \mathrm{Set}$ you see that you only need a non-identity map of right M-Sets between $X$ and itself ($X$ is the image of the only object in the category $\mathrm{M}$), which exists in general.

  • 0
    Thank you, pleaae see my comment above2012-06-03