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# Exercise 95 - Chapter 3

edited June 2018

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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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.
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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}\$$.