2
$\begingroup$

Suppose $F: C \rightarrow D$ is the left adjoint to a functor $G$. Then is it true that the functor $F^{\star}:[C : Sets]$ defined by prescomposing a functor $P: C \rightarrow Sets$ is still left adjoint to the functor $G^{\star}$ defined similarly?

Or would the adjunction be flipped an $F^{\star}$ becomes right adjoint to $G^{\star}$?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ This is an immediate consequence of the fact that $[-, \mathbf{Set}] : \mathfrak{Cat}^\mathrm{op} \to \mathfrak{CAT}$ is a 2-functor; note that the handedness of the adjunction is reversed. Amusingly, $[(-)^\mathrm{op}, \mathbf{Set}] : \mathfrak{Cat}^\mathrm{coop} \to \mathfrak{CAT}$ preserves the handedness of adjunctions. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2014 at 10:45

1 Answer 1

Reset to default
3
$\begingroup$

Yes,

Let $Q \in [\mathfrak{D}:Sets]$,$A \in \mathfrak{D}$ $P \in [\mathfrak{C}:Sets]$ and $B \in \mathfrak{C}$.

If $F: \mathfrak{C} \rightarrow \mathfrak{D}$ is left adjoint to $G$, then there are unit and counit natural transformations $\eta: F \circ G \rightarrow 1_{\mathfrak{D}}$ and $\epsilon: G\circ F \rightarrow 1_{\mathfrak{C}}$ respectively.

These induce natural ismorphisms $\epsilon^{\star}: G^{\star}\circ F^{\star} (Q)(A) = Q ((F\circ G) (A)) \rightarrow Q(1_{\mathfrak{D}}(A)) = Q(A) \cong 1_{[\mathfrak{D}:Sets]} (Q(A))$. Therefore, the counit has induces a unit.

Dually, we have $\eta^{\star}: F^{\star}\circ G^{\star} (P)(B) = P ((G\circ F) (B)) \rightarrow P(1_{\mathfrak{C}}(B)) = P(B) \cong 1_{[\mathfrak{C}:Sets]} (P(B))$ and similarly, the unit has induced a counit.

Therefore if $F$ is left adjoint to $G$, then $G^{\star}$ is left adjoint to $F^{\star}$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .