I'm trying to understand the concept of adjunction in category theory.

This wikipedia article says that if $F$ is a functor from $\mathcal{D}$ to $\mathcal{C}$ and $G$ is a functor from $\mathcal{C}$ to $\mathcal{D}$, then we say $F$ is left-adjoint (or $G$ is right-adjoint) when:

there is a bijection between $\hom_{\mathcal{C}}(F(Y), X)$ and $\hom_{\mathcal{D}}(Y, G(X))$, which is natural for every objects $X\in\mathcal{C}$ and $Y\in\mathcal{D}$.

I'm trying to figure out what "natural" means in this context.

I'm familiar with the notion of natural transformation $\eta$ between functors $F_1$ and $F_2$ in which for every object $F_1(A)$ there is a morphism $\eta_A:F_1(A)\to F_2(A)$ such that if $f:A\to B$, then $\eta_B\circ F_1(f)=F_2(f)\circ \eta_A$.

But I don't know how to apply it in this context.

Any ideas? Thanks!

  • 2
    $\begingroup$ That's the kind of "natural" it means; it's a natural transformation. In this case between $\hom(F(-),-)$ and $\hom(-,G(-))$. $\endgroup$ Mar 24, 2017 at 18:04
  • 2
    $\begingroup$ More formally it means that there is a natural isomorphism between the functors $ Hom_{C}(F(\cdot), \cdot) \to Hom_{D}(\cdot, G(\cdot))$ which are both functors from $C^{op}\times D \to Set$ $\endgroup$ Mar 24, 2017 at 18:04
  • $\begingroup$ @Max, that was really helpful, thank you $\endgroup$
    – rmdmc89
    Mar 24, 2017 at 21:44
  • $\begingroup$ But @Max, it is not clear to me what the functor $Hom_C(F(\cdot), \cdot)$ does with a morphism in category $C^{op}\times D$. It should give us a function, but what function? $\endgroup$
    – rmdmc89
    Mar 24, 2017 at 21:53
  • 1
    $\begingroup$ @Aguirrek : if you have an arrow $(f^{op}, g) : (X,Y) \to (T,Z)$ in $C^{op}\times D$, then you actually have arrows $f: T\to X$ and $g: Y\to Z$. To these, you send $\phi : Hom_C(F(X), Y) \to Hom_C(F(T), Z)$ defined by $\phi(h) = g\circ h\circ F(f)$. You can check that it does make $Hom_C(F(\cdot), \cdot)$ a functor, but that's just a bit of diagram chasing $\endgroup$ Mar 24, 2017 at 22:18

3 Answers 3


For completeness, the point is that you can define two functors

$$ U,V : \mathcal{D}^\text{op} \times \mathcal{C} \to \mathbf{Set} $$

by the formulas

$$ U(d,c) = \hom_\mathcal{C}(F(d), c) $$ $$ V(d,c) = \hom_\mathcal{D}(d, G(c)) $$

and these two functors should be naturally isomorphic.

If you're not familiar with the hom-functor, for any category $\mathcal{C}$, $\hom$ is a functor

$$ \mathcal{C}^\text{op} \times \mathcal{C} \to \mathbf{Set} $$

On objects $X$ and $Y$, $\hom(X,Y)$ is the set of morphisms from $X$ to $Y$.

For an object $X$ and morphism $f : Y \to Z$, you have

$$ \hom(X,f) : \hom(X,Y) \to \hom(X,Z) : g \mapsto f \circ g $$

For a morphism $h : W \to X$ and an object $Y$,

$$ \hom(h,Y) : \hom(X,Y) \to \hom(W,Y) : g \mapsto g \circ h $$

And for a pair of morphisms $h : W \to X$ and $f : Y \to Z$,

$$ \hom(h,f) : \hom(X,Y) \to \hom(W,Z) $$

is simply the composite

$$ \hom(h,f) = \hom(h,Z) \circ \hom(X, f) = \hom(W,f) \circ \hom(h,Y)$$

or more simply,

$$ g \mapsto f \circ g \circ h $$


I'm going to assume that $G$ is covariant, but similar statements hold regardless of what types of functors each is. If you have a morphism $X\to X'$ in $\mathcal C$, then you have a morphism $G(X)\to G(X')$ in $\cal D$, and we say it's natural in $X$ if for each $Y\in\cal D$ the following diagram commutes:

$$\begin{array}{cccc} \hom_{\cal C}(F(Y),X) & \xrightarrow{} & \hom_{\cal D}(Y,G(X)) \\ \downarrow & & \downarrow \\ \hom_{\cal C}(F(Y),X') & \xrightarrow{} & \hom_{\cal D}(Y,G(X')) \end{array}$$

and similarly, it's natural in $Y$ if for each morphism $Y\to Y'$ in $\cal D$ the analogous diagram commutes.


The most concrete definition is as an initial object in a comma category. Even more concretely phrased, it's "like the free group functor $F: \mathbf{Set} \to \mathbf{Grp}$ with the forgetful functor $U: \mathbf{Grp} \to \mathbf{Set}$".

The free group functor takes a set $X$ to its free group $FX$, and a function $f: X \to Y$ to the "do the function $f$ pointwise to each letter of the word" homomorphism $FX \to FY$.

The forgetful functor just takes a group and views its underlying set, and it takes a group homomorphism and calls it a function instead.

Let $X$ be a set and $G$ a group. Then there's an inclusion function $\iota_X: X \to UFX$ sending $x$ to the word $(x)$ of length $1$. (Recall that $UFX$ is the set of all words on the letters in $X$.)

Then the statement that $U$ and $F$ are adjoint is the same as the statement that for every function $f: X \to UG$ there is a unique group homomorphism $g: FX \to G$ such that $f = (Ug) \circ \iota_X$.

Hopefully you can convince yourself that this fact is actually true of groups and sets by contemplating it with a cup of hot chocolate. This is how I prefer to think of the adjoint, because it's so concrete in a world where everything is alarmingly abstract.


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