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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 $?

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    $\begingroup$ Let $A = B = \mathbb{B} G$, where $G$ is a group and $\mathbb{B} G$ is $G$ considered as a one-object category. Let $F$ be the identity functor. Then there is exactly one natural transformation of $F$ for each element of the centre of $G$. $\endgroup$
    – Zhen Lin
    Jun 3, 2012 at 11:43
  • $\begingroup$ @ZhenLin and tetrapharmakon and Andy : thank you for your quick answers. I came to the same conclusions and similar examples over lunch, right after asking the question. I am trying to gain some insight into functor categories in general, so I will shortly place a related but perhaps more interesting/challenging question. $\endgroup$
    – magma
    Jun 3, 2012 at 13:53
  • $\begingroup$ @ZhenLin just a notational curiosity: why did you write BG ? what does the double strike B stand for? Is it standard notation? $\endgroup$
    – magma
    Jun 3, 2012 at 14:00
  • $\begingroup$ 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 $[\mathbb{B} G^\textrm{op}, \textbf{Set}]$. $\endgroup$
    – Zhen Lin
    Jun 3, 2012 at 16:38

2 Answers 2

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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).

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  • $\begingroup$ I'm never bothered by generality :) $\endgroup$
    – Andy
    Jun 3, 2012 at 11:54
  • $\begingroup$ Thank you, pleaae see my comment above $\endgroup$
    – magma
    Jun 3, 2012 at 13:54
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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.

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  • $\begingroup$ Thank you, pleaae see my comment above $\endgroup$
    – magma
    Jun 3, 2012 at 13:54

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