Exercise 3.27

Question

Show that any (covariant or contravariant) functor from $\mathrm{C}$ to $\mathrm{D}$ takes isomorphisms in $\mathrm{C}$ to isomorphisms in $\mathrm{D}$.

Elu Thingol · Accepted Answer

Suppose first that $\mathcal{F}$ is a covariant functor from $\mathrm{C}$ to $\mathrm{D}$, and let $f \in \operatorname{hom}_{\mathrm{C}}(X, Y)$ be an isomorphism. We have
$$\operatorname{Id}_{\mathcal{F}(X)}=\mathcal{F}\left(f^{-1} \circ f\right)=\mathcal{F}\left(f^{-1}\right) \circ \mathcal{F}(f)$$

$$\operatorname{Id}_{\mathcal{F}(Y)}=\mathcal{F}\left(f \circ f^{-1}\right)=\mathcal{F}(f) \circ \mathcal{F}\left(f^{-1}\right).$$

In other words, $\mathcal{F}(f)$ is an isomorphism, with inverse $\mathcal{F}\left(f^{-1}\right)$.

In the case that $\mathcal{F}$ is a contravariant functor, the proof follows analogously.

Exercise 3.27

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Exercise 3.27

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