Image of a natural transformation under an equivalence

Question

Let $C,D$ be a categories, $F,G:C\times C\to C$ functors with an natural isomorphism $\eta_X$ between them, and $E:C\to D$ an equivalence.

Is it true that $E(\eta_X)$ is a natural isomorphism as well?

Answer 1

Given any morphism $f: (A,B) \to (A',B')$ in $C \times C$, then with your natural transformation $\eta_{(A,B)}: F(A,B) \to G(A,B)$ you have $$G(f) \circ \eta_{(A,B)} = \eta_{(A',B')} \circ F(f)$$ Applying $E$ on both sides, you get $$EG(f) \circ E(\eta_{(A,B)}) = E(\eta_{(A',B')}) \circ EF(f)$$ so $\{E(\eta_{(A,B)})\}_{(A,B)}$ is a natural transformation between $EF$ and $EG$.

Note that $\eta$ is an isomorphism iff $\eta_{(A,B)}$ is an isomorphism for all objects $(A,B)$ in $C \times C$. Since functors map isomorphisms to isomorphisms we find that $E(\eta)$ is a natural isomorphism if $\eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C \to D$.

Answer 2

Using juxtaposition for horizontal composition and $\cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)

Let $\eta : F \to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,

$$(E \eta) \cdot (E \eta^{-1}) = (E \cdot E) (\eta \cdot \eta^{-1}) = E G $$ $$(E \eta^{-1}) \cdot (E \eta) = (E \cdot E) (\eta^{-1} \cdot \eta) = E F $$

Both vertical composites are the respective identity natural transformations, and thus $E \eta^{-1}$ is the inverse of $E \eta$.

Answer 3

I think as a matter of notation, it's worth asking what $E(\eta_X)$ really means—it's just a shorthand for the horizontal composition $\text{id}_{E}\circ \eta_X\in \text{Hom}_{D^{C\times C}}(E\circ F,E\circ G)$. And seeing as $\text{id}_{E}$, $\eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $\text{id}_E^{-1}\circ \eta_X^{-1}$).