Understanding a proof of a lemma for rigid categories

Question

I'm reading the proof of the lemma 3.4 in the Bruguieres' paper on Hopf monads which claims the following:

Lemma Let $F,G: \mathcal{C} \rightarrow \mathcal{D}$ be two strong monoidal functors and $\alpha: F\Rightarrow G$ a monoidal natural transformation. If $\mathcal{C}$ is left (or right) rigid, then $\alpha$ is a natural isomorphism.

In the proof, they claim the following: $${}^{\vee}\alpha_Y\alpha_X=(e_G(\alpha_X\otimes \alpha_Y)\otimes \mathrm{id}_{F(X)})(\mathrm{id}_{F(X)}\otimes d_F),$$ and $$\alpha_X{}^{\vee}\alpha_Y=(e_G\otimes \mathrm{id}_{G(X)})(\mathrm{id}_{G(X)}\otimes(\alpha_Y\otimes \alpha_X))d_F)$$

where $Y$ is the left (resp. right) dual of $X$ and $e_G, d_F$ denote evaluation and coevaluation morphisms obtained from $F,G$.

I try to understand why, but I can't see it.

Answer 1

Notation:

1. Let $F_2: F(X\otimes Y) \xrightarrow{\sim} F(X)\otimes F(Y)$ and $F_0: F(\mathbb{1}) \xrightarrow{\sim} \mathbb{1}$ denote the monoidality isomorphisms which exhibit $F:\mathcal{C}\rightarrow \mathcal{D}$ as a strong monoidal functor. Similarly, denote by $G_2, G_0$ the corresponding structure isomorphisms for $G: \mathcal{C}\rightarrow \mathcal{D}$.
2. Let $Y = {}^\vee X$ be the left dual of $X$ in $\mathcal{C}$ with evaluation $\mathrm{ev}: X\otimes Y\rightarrow \mathbb{1}$ and coevaluation $\mathrm{coev}: \mathbb{1}\rightarrow Y\otimes X$. Since $F$ and $G$ are monoidal they induce a duality on $F(X)$ resp. $G(X)$ as follows: The dual of $F(X)$ is $F(Y)$ with evaluation $$e_F = F_0\circ F(\mathrm{ev})\circ F_2^{-1}: F(X)\otimes F(Y)\rightarrow \mathbb{1} $$ and coevaluation $$ d_F: F_2\circ F(\mathrm{coev}) F_0^{-1}: \mathbb{1} \rightarrow F(Y)\otimes F(X)$$ and similarly for $G$.

Answer:

The only non-trivial step in your question is to show the identity: $$e_G\circ (\alpha_X \otimes \alpha_Y) = e_F.$$ To prove this first use monoidality of $\alpha$: $$e_F\circ (\alpha_X\otimes \alpha_Y) = G_0\circ G(\mathrm{ev})\circ G_2^{-1}\circ (\alpha_{X}\otimes \alpha_Y) = G_0\circ G(\mathrm{ev})\circ \alpha_{X\otimes Y} \circ F_{2}^{-1}$$ and now that $\alpha$ is natural: $$G_0\circ G(\mathrm{ev})\circ \alpha_{X\otimes Y} \circ F_{2}^{-1} = G_0 \circ \alpha_{\mathbb{1}} \circ F(ev)\circ F_{2}^{-1} = e_F.$$ In the last step we used again that $\alpha$ is monoidal.

(The second equation follows similarly.)