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Exercise 95 - Chapter 3

Recall from Definition 3.26 that every category \(\mathcal{C}\) has an opposite \(\mathcal{C}^{op}\). Let \(F : \mathcal{C} \rightarrow \mathcal{D} \) be a functor. How should we define its opposite, \(F^{op} : \mathcal{C}^{op} \rightarrow \mathcal{D}^{op} \)? That is, how should \( F^{op} \) act on objects, and how should it act on morphisms?

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Comments

  • 1.

    Huh. I mean the Functor \(F' : C^{OP} \to D^{Op}\) is straight forward but I'm not sure \(F^OP\) as defined above always exists.

    Comment Source: Huh. I mean the Functor \\(F' : C^{OP} \to D^{Op}\\) is straight forward but I'm not sure \\(F^OP\\) as defined above always exists.
  • 2.

    \(F^\mathrm{op}\) should do the same thing as \(F\) on objects (i.e., \(\forall c\in\mathcal{C}\,\,F(c)=F^\mathrm{op}(c)\)). On morphisms, I imagine it should take \(f^\mathrm{op}\) to \((F(f))^\mathrm{op}\). This works trivially on identities. \(F^\mathrm{op}(f^\mathrm{op}\circ g^\mathrm{op})=F^\mathrm{op}((g\circ f)^\mathrm{op})\) \(=(F(g\circ f))^\mathrm{op}=(F(g)\circ F(f))^\mathrm{op}\) \(=(F(f))^\mathrm{op}\circ(F(g))^\mathrm{op}\) \(=F^\mathrm{op}(f)\circ F^\mathrm{op}(g)\), so \(F^\mathrm{op}\) defined in this way is a functo from \(\mathcal{C}^\mathrm{op}\) to \(\mathcal{D}^\mathrm{op}\).

    Comment Source:\\(F^\mathrm{op}\\) should do the same thing as \\(F\\) on objects (i.e., \\(\forall c\in\mathcal{C}\,\,F(c)=F^\mathrm{op}(c)\\)). On morphisms, I imagine it should take \\(f^\mathrm{op}\\) to \\((F(f))^\mathrm{op}\\). This works trivially on identities. \\(F^\mathrm{op}(f^\mathrm{op}\circ g^\mathrm{op})=F^\mathrm{op}((g\circ f)^\mathrm{op})\\) \\(=(F(g\circ f))^\mathrm{op}=(F(g)\circ F(f))^\mathrm{op}\\) \\(=(F(f))^\mathrm{op}\circ(F(g))^\mathrm{op}\\) \\(=F^\mathrm{op}(f)\circ F^\mathrm{op}(g)\\), so \\(F^\mathrm{op}\\) defined in this way is a functo from \\(\mathcal{C}^\mathrm{op}\\) to \\(\mathcal{D}^\mathrm{op}\\).
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