Left (resp. right) adjoint functor fully faithful iff unit (resp. counit) isomorphism

741 Views Asked by At

Let $\mathcal{C}$ and $\mathcal{D}$ be categories and $\mathcal{F}:\mathcal{C}\longleftrightarrow\mathcal{D}:G$ functors such that $\mathcal{G}$ is right-adjoint to $\mathcal{F}$, ie. we have a natural isomorphism $\text{Hom}_{\mathcal{D}}(\mathcal{F}(X),Y)\cong \text{Hom}_{\mathcal{C}}(X,\mathcal{G}(Y))$ for all objects $X$ in $\mathcal{C}$ and $Y$ in $\mathcal{D}$. I have proven that right-adjoint functors are unique up to natural transformation and I have constructed the unit $\eta:\text{Id}_{\mathcal{C}}\longrightarrow G\circ F$ as well as the counit $\varepsilon:F\circ G\longrightarrow\text{Id}_{\mathcal{D}}$. But now I am stuck: How can I show that $\mathcal{F}$ (resp. $\mathcal{G}$) is fully faithful if and only if $\eta$ (resp. $\varepsilon$) is an isomorphism?

1

There are 1 best solutions below

0
On

For any pair of objects $X,X' \in \mathcal C$, we can consider the induced map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$. Via naturality, we know that this composite map is actually equal to $f \mapsto \eta_{X'} \circ f$. This is just $\mathrm{Hom}_{\mathcal C}(-,\eta_{X'})$. From the Yoneda lemma, we know that this is a bijection for all $X$ iff $\eta_{X'}$ is an isomorphism.

On the other hand, $\mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$ is always a bijection, since we have an adjunction, so that the composite map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$ is a bijection iff the map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X'))$ is a bijection. This is precisely the condition that $\mathcal F$ is fully faithful.

The statement for $\varepsilon$ is dual to the one about $\eta$.