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Let $(L,R)$ be a pair of adjoint functor.

How to show that the commutativity of the left diagram induces the commutativity of the right one?

diagram

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Let $(\eta,\epsilon):F \dashv G:\mathfrak{C} \to \mathfrak{D}$ be an adjoint pair of functors and assume you have the commuting diagram $$ \array{ A & \xrightarrow{f} & GX \\ h \downarrow & & \downarrow G(k) \\ B & \xrightarrow{g} & GY } $$ in $\mathfrak{C}$. Then applying the functor $F$ to the diagram gives the commuting diagram $$ \array{ FA & \xrightarrow{F(f)} & F(GX) \\ F(h) \downarrow & & \downarrow F(G(k)) \\ FB & \xrightarrow{F(g)} & F(GY) } $$ in $\mathfrak{D}$. Using the counit allows us to produce the commuting diagram $$ \array{ FA & \xrightarrow{F(f)} & F(GX) & \xrightarrow{\epsilon_{X}} & X \\ F(h) \downarrow & & \downarrow F(G(k)) & & \downarrow k\\ FB & \xrightarrow{F(g)} & F(GY) & \xrightarrow{\epsilon_Y} & Y } $$ which contracts to the commuting diagram $$ \array{ FA & \xrightarrow{\epsilon_X \circ F(f)} & X \\ F(h) \downarrow & & \downarrow k \\ FB & \xrightarrow{\epsilon_Y \circ F(g)} & Y } $$ in $\mathfrak{D}$. There is some work to check that the middle rectangle actually commutes, but it all follows from the triangle identities of adjunction and the couniversal property of $\epsilon$.

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    $\begingroup$ I disagree that you need the triangle identities here. You applied $F$, so the left hand square commutes, while the right hand square commutes by naturality of $\epsilon$. $\endgroup$ Feb 5 '17 at 22:48
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You probably know that $L : \mathfrak C \to \mathfrak D$ is left adjoint to $R : \mathfrak D \to \mathfrak C$ if an only if it exists a natural bijection $$ \mathfrak C (A,R(X)) \stackrel {\alpha_{A,X}} {\overset\sim\to} \mathfrak D (L(A), X). $$

Now what does it mean that this $\alpha$ is natural? It is a transformation between the two functors $\mathfrak C^{\rm op} \times \mathfrak D \to \mathbf{Set}$. In particular, you should check what naturality gives you for arrows of $\mathfrak C^{\rm op}\times \mathfrak D$ of the form $(f,{\rm id})$ and $({\rm id},g)$.

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Suppose $L : \mathcal{C} \to \mathcal{D}$ is left adjoint to $R : \mathcal{D} \to \mathcal{C}$, with unit $\eta : 1_\mathcal{C} \to RL$ and counit $\epsilon : LR \to 1_\mathcal{D}$.

We can prove the following, more general statement: let $J$ be a category, and let $[J, X]$ be the category of functors from $J$ to $X$. Let $L_*$ denote the composition functor $[J, \mathcal{C}] \to [J, \mathcal{D}]$, and similarly for $R_*$.

Theorem: $L_*$ is left adjoint to $R_*$

The unit $\eta_*$ and counit $\epsilon_*$ are given by composition as well: e.g. given a functor $F : J \to \mathcal{C}$, the component $ \eta_{*F} : F \to RLF$ is the horizontal composite $\eta F$.

That the triangle identities hold for $\eta_*$ and $\epsilon_*$ follows nearly immediately from the fact they hold for $\eta$ and $\epsilon$.

The whole thing can sort of be interpreted as taking the horizontal composite of an adjunction with a functor or natural transformation.


The given problem is the case where $J$ is the arrow category $\bullet \to \bullet$, so that a natural transformation between two functors from $J$ is a commutative square. Let $F$ be the diagram $ A \to B$ and $G$ be the diagram $X \to Y$. Then, we have a natural isomorphism

$$\hom_{[J, \mathcal{D}]}(LF, G) \cong \hom_{[J, \mathcal{C}]}(F, RG) $$

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