Natural transformations and composition of functors

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text public domain (CC0).

A functor compares objects and arrows in two categories. A natural transformation compares two such comparisons. Its components cannot be chosen independently: every arrow in the source imposes an equation. These equations let us compose transformations, change the category in which a diagram takes values, and compare several successive functors without losing their compatibility.

We assume the category, functor and size conventions in Universes and small categories, and the definition of a faithful or fully faithful functor in Relations and cancellation in categories. All object and arrow collections are sets in a sufficiently large ambient universe. Local smallness refers to the specified smaller universe. The last example uses elementary vector spaces over any field. Basic references are [Stacks, Categories], [Schapira, Categories and Homological Algebra] and the preceding two lessons. We use the exact open horizontal-composition proofs in [Stacks], with their notation explained below.

1. Maps between diagrams form a category

Let \(F,G:\mathcal C\to\mathcal D\) be functors. A natural transformation \(\alpha:F\Rightarrow G\) consists of arrows \(\alpha_X:F(X)\to G(X)\), one for each object \(X\) of \(\mathcal C\), satisfying

The component \(\alpha_X\) is also written \(\alpha(X)\). Equation (1.1) says that the two routes from \(F(X)\) to \(G(Y)\) agree. It involves every source arrow, not only the isomorphisms. We write composites from right to left.

The identity transformation \(1_F\) has component \(1_{F(X)}\). It is natural because both sides of (1.1) are \(F(f)\). If \(\alpha:F\Rightarrow G\) and \(\beta:G\Rightarrow H\), define their vertical composite by

Proposition 1.1. Functors \(\mathcal C\to\mathcal D\), with natural transformations as arrows, form a category \(\operatorname{Fun}(\mathcal C,\mathcal D)\).

Proof. The proposed composite has source \(F(X)\) and target \(H(X)\). For \(f:X\to Y\), the two naturality equations give

Thus the composite is natural. A third transformation \(\gamma:H\Rightarrow K\) satisfies \(((\gamma\beta)\alpha)_X=\gamma_X(\beta_X\alpha_X)=(\gamma(\beta\alpha))_X\), by associativity in \(\mathcal D\). Equality at every component is equality of transformations. Likewise \((1_G\alpha)_X=\alpha_X=(\alpha1_F)_X\). These are associativity and both identity laws.

Functor data are a subset of pairs of functions on the ambient object and arrow sets, subject to the source, target, identity and composition equations. They therefore form an ambient set. For fixed \(F,G\), transformation data form a subset of the ambient product of the sets \(\operatorname{Hom}_{\mathcal D}(F(X),G(X))\); naturality selects a further subset. Thus all the objects and Hom sets used here are legitimate in the ambient universe. \(\square\)

The notation \(\mathcal D^{\mathcal C}\) also means this functor category. Its objects are functors; a Hom set consists of natural transformations.

For local smallness, we retain the complete proof of Universes and small categories, Proposition 5.1. If \(\operatorname{Ob}(\mathcal C)\) is \(\mathcal U\)-small and \(\mathcal D\) is locally \(\mathcal U\)-small, then every \(\operatorname{Nat}(F,G)\) is \(\mathcal U\)-small. The source Hom sets may be larger; the target object set need not be small. In particular this applies when \(\mathcal C\) is a small category. The empty source is included: there is one empty functor and one empty transformation. This does not make the object set of a general functor category \(\mathcal U\)-small.

2. Change the values of a diagram

Let \(P:\mathcal D\to\mathcal E\) be a functor. Postcomposition sends \(F:\mathcal C\to\mathcal D\) to \(PF\), and sends \(\alpha:F\Rightarrow G\) to the family

Applying \(P\) to (1.1) gives \(PG(f)P(\alpha_X)=P(\alpha_Y)PF(f)\), so the family is natural. The identities \(P(1_{F(X)})=1_{PF(X)}\) and \(P(\beta_X\alpha_X)=P(\beta_X)P(\alpha_X)\) prove that this is a functor

There is also precomposition. For \(Q:\mathcal B\to\mathcal C\), send \(F\) to \(FQ\) and \(\alpha\) to \(\alpha Q\), with component \(\alpha_{Q(B)}\). Naturality at \(b:B\to B'\) is exactly (1.1) at \(Q(b)\). Componentwise identities and composites are preserved. These operations are also called left and right whiskering, respectively.

Lemma 2.1. A natural transformation is an isomorphism in its functor category if and only if every component is an isomorphism.

Proof. If \(\beta\) is an inverse to \(\alpha\), evaluating both inverse equations at \(X\) makes \(\beta_X\) an inverse to \(\alpha_X\). Conversely, assume every \(\alpha_X\) has an inverse and put \(\beta_X=\alpha_X^{-1}\). From \(G(f)\alpha_X=\alpha_YF(f)\), compose on the left with \(\alpha_Y^{-1}\) and on the right with \(\alpha_X^{-1}\). This gives

So \(\beta:G\Rightarrow F\) is natural. Both composites have identity components and hence are identity transformations. Inverses are unique by the ordinary two-sided inverse argument in Relations and cancellation. The empty source again satisfies the statement. \(\square\)

Theorem 2.2. If \(P\) is faithful, so is \(P_*\). If \(P\) is fully faithful, so is \(P_*\).

Proof. Suppose \(P\alpha=P\beta\) for two transformations with the same source and target. For every \(X\), faithfulness of \(P\) gives \(\alpha_X=\beta_X\). Thus \(\alpha=\beta\).

For fullness, let \(\tau:PF\Rightarrow PG\). Full faithfulness of \(P\) gives a unique arrow \(\alpha_X:F(X)\to G(X)\) with \(P(\alpha_X)=\tau_X\), for each \(X\). The naturality equation of \(\tau\) is

These are images of parallel arrows in \(\mathcal D\). Faithfulness reflects their equality, proving that \(\alpha\) is natural. Its components give \(P\alpha=\tau\), and their uniqueness makes the lift unique. With empty \(\mathcal C\), there are no components to choose and both Hom sets are singletons, so the same argument holds. \(\square\)

Fullness supplies components; faithfulness supplies their compatibility. This is why the second proof uses both conditions.

3. Compose comparisons along successive functors

Take functors \(F_0,F_1:\mathcal C\to\mathcal D\) and \(G_0,G_1:\mathcal D\to\mathcal E\), with transformations \(\alpha:F_0\Rightarrow F_1\) and \(\beta:G_0\Rightarrow G_1\). Whiskering and vertical composition give two transformations from \(G_0F_0\) to \(G_1F_1\). Their equality is the horizontal two-path identity:

We use the complete open proof in Stacks, the component square preceding horizontal composition. Its \(t\) is \(\alpha\), its \(s\) is \(\beta\), and its \(x\) is \(X\); its categories \(\mathcal A,\mathcal B,\mathcal C\) are our \(\mathcal C,\mathcal D,\mathcal E\). The square has upper left object \(G_0F_0(X)\) and lower right object \(G_1F_1(X)\), precisely the endpoints in (3.1). There is no size, additivity or invertibility hypothesis in that proof.

The transformation defined by (3.1) is the horizontal composite \(\beta*\alpha\). It is natural: its first expression is the vertical composite of the two natural whiskerings \(G_0\alpha\) and \(\beta F_1\), established in Section 2. The second expression is the other possible order of changing the functors. The formula also gives

Indeed the additional component in either expression is an identity. In particular \(1_G*1_F=1_{GF}\).

Theorem 3.1. Horizontal composition is associative, has the identity transformations of identity functors as units, and obeys the interchange law

where \(\alpha:F_0\Rightarrow F_1\), \(\alpha':F_1\Rightarrow F_2\), \(\beta:G_0\Rightarrow G_1\) and \(\beta':G_1\Rightarrow G_2\) have the indicated successive types.

Proof. The complete interchange proof is Stacks, the proof of the horizontal and vertical composition properties. Substitute \(t=\alpha\), \(t'=\alpha'\), \(s=\beta\) and \(s'=\beta'\). Vertical composition there is \(\circ\), and horizontal composition is \(\star\). Its equation is exactly (3.3), with source \(G_0F_0\) and target \(G_2F_2\). This proof uses the two-path equality already identified above and functoriality; it applies to arbitrary transformations, including noninvertible ones.

Here are the associativity and unit calculations. Add functors \(H_0,H_1:\mathcal E\to\mathcal K\) and \(\gamma:H_0\Rightarrow H_1\). Both \((\gamma*\beta)*\alpha\) and \(\gamma*(\beta*\alpha)\) have component

For the first composite this follows by applying (3.1) first to \(\gamma*\beta\); for the second, apply (3.1) to \(\beta*\alpha\), then use that \(H_0\) preserves its composite. Associativity in \(\mathcal K\) makes the parenthesizations of (3.4) equal. Thus the transformations agree. Ordinary functor composition is itself literally associative on objects and arrows, so their sources and targets are the same functors.

For the left unit, \(1_{\operatorname{id}_{\mathcal D}}*\alpha\) has component \(1_{F_1(X)}\operatorname{id}_{\mathcal D}(\alpha_X)=\alpha_X\). For the right unit, \(\alpha*1_{\operatorname{id}_{\mathcal C}}\) has component \(\alpha_XF_0(1_X)=\alpha_X\). The identity functors also preserve the source and target functors literally. This proves both horizontal unit laws. The vertical unit and associativity laws are Proposition 1.1. \(\square\)

Consequently there is a functor

which sends \((\beta,\alpha)\) to \(\beta*\alpha\). The arrow in the product category is a pair of transformations. Its identity is a pair of identities, preserved by (3.2); its composite is componentwise vertical composition, preserved by (3.3). This proves bifunctoriality on arrows, as well as the object assignment.

One can keep track of all three levels at once. Take small \(\mathcal U\)-categories with data in a fixed larger universe as objects, functors as arrows, and natural transformations as arrows between functors. The category-data encoding is in Universes and small categories, Section 4. The collection of those data is a set in a further ambient universe. Functors between two objects form an ambient set by Proposition 1.1. Identity assignments and composition of functors give ordinary identities and composition; the complete composite-functor calculation is in Relations and cancellation in categories, Section 3. Associativity and both unit laws hold on the object and arrow functions. This constructs the ordinary category of small categories and functors within the stated size bounds.

Its ordinary Hom set has functors as elements. Equipping that set of objects with natural transformations gives the Hom category \(\operatorname{Fun}(\mathcal C,\mathcal D)\). It is the set of objects of this Hom category, rather than the entire category, that is the ordinary Hom set.

A strict 2-category has these Hom categories, functorial composition between them, and identity arrows, with associative and unit laws holding as equalities. Proposition 1.1, (3.5), Theorem 3.1 and strict ordinary functor composition verify every one of these requirements here. “Strict” refers to those equalities; it does not say that every natural transformation is invertible. No claim that the collection of all small categories is itself \(\mathcal U\)-small is required.

4. The identity functor has a commutative center

Write

Its elements are families \(z_X:X\to X\) such that \(fz_X=z_Yf\) for every \(f:X\to Y\). Vertical composition and the identity transformation make it a monoid, by Proposition 1.1. This monoid is commutative. We use the complete component-naturality argument in Ring actions and tensor–Hom in abelian categories, Section 1: its \(\eta,\theta\) are any two elements of (4.1). That calculation uses only naturality for the arrow \(\theta_X:X\to X\), so it applies here without the additive hypotheses needed for the ring structure in that lesson. Thus \(z_Xw_X=w_Xz_X\) for every \(X\), which is exactly equality of the two composite transformations.

The subset \(\operatorname{Aut}(\operatorname{id}_{\mathcal C})\) consists of the units of this monoid. It is a group: identities and composites are invertible; if \(z,w\) have inverses, \(w^{-1}z^{-1}\) is a two-sided inverse of \(zw\). Lemma 2.1 makes every componentwise inverse natural. Each inverse therefore lies in the same subset. The group is abelian because its multiplication is the restriction of the commutative monoid law. For the empty category, both the monoid and the group are singletons.

Natural isomorphisms also explain the phrase quasi-commutative diagram. When a diagram has categories at its vertices and functors on its edges, two indicated composite routes may be naturally isomorphic without being equal. A specified natural isomorphism gives their comparison. An isomorphism of categories, in contrast, has an inverse functor whose two composites are literally the identity on objects and arrows.

Comparisons for several routes can require further equations. To see this concretely, take the one-object category whose arrows form the two-element group \(\{1,u\}\), with \(u^2=1\). Let three route functors \(P,Q,R\) all be its identity functor. Both \(1\) and \(u\) give natural automorphisms, since this group is commutative. Choose the comparisons \(P\Rightarrow Q\) and \(Q\Rightarrow R\) to have component \(1\), but choose \(P\Rightarrow R\) to have component \(u\). Every comparison is an isomorphism, yet the composite of the first two differs from the third. The existence of route comparisons alone therefore imposes no automatic compatibility among them.

5. Evaluation is natural before it is invertible

Let \(k\) be any field. Fix a universe containing \(k\), and consider vector spaces whose underlying sets belong to it. For any such \(V\), its algebraic dual is \(V^*=\operatorname{Hom}_k(V,k)\). Addition and scalar multiplication are pointwise. For \(a,b,c,d\in k\) and \(v,w\in V\), their results remain linear functionals because

The zero functional belongs to this set. The vector-space identities hold after evaluation at every \(v\), where they are the identities in \(k\). This proves the vector-space structure for all \(V\).

For a linear map \(f:V\to W\), its dual is

The composite is a linear functional. The equality \((a\lambda+b\mu)f=a(\lambda f)+b(\mu f)\), evaluated on every \(v\), proves that \(f^*\) is linear. For \(g:W\to T\), evaluation on a functional \(\nu\in T^*\) gives \((gf)^*(\nu)=\nu gf=f^*g^*(\nu)\); the dual of an identity is an identity. Thus dualization is a functor \(\mathsf{Vect}_k^{\mathrm{op}}\to\mathsf{Vect}_k\). Applying it twice gives a covariant functor: \((gf)^{**}=g^{**}f^{**}\), and identities are preserved twice.

The evaluation maps are the maps \(\delta_V\) in Finite duality and failure of opposite density, Section 1:

For fixed \(v\), the right side is linear in \(\lambda\), so it is an element of \(V^{**}\). The equalities \(\lambda(av+bw)=a\lambda(v)+b\lambda(w)\), for all \(\lambda\), make \(\delta_V\) linear in \(v\). The all-vector-space evaluation equality \(f^{**}\delta_V=\delta_Wf\) from the cited lesson has the following explicit component meaning:

Equality for every \(\lambda\in W^*\) is equality in \(W^{**}\). Hence \(\delta:\operatorname{id}\Rightarrow(-)^{**}\) is a natural transformation on all vector spaces. No basis, finite-dimensionality or isomorphism hypothesis is used for naturality. The finite-dimensional isomorphism and the behavior of the double dual in infinite dimension are separate questions treated in the cited lesson.

6. Exercises with full solutions

Exercise 1 (beginner: all maps between two short diagrams). Let \([1]\) be the category with objects \(0,1\), their identities and one arrow \(s:0\to1\). Define set-valued functors by

List every natural transformation \(F\Rightarrow G\). How many arbitrary component families are there before naturality is imposed? Is any resulting transformation an isomorphism?

Solution. A component at \(0\) chooses either \(c\) or \(d\). A component at \(1\) is an arbitrary function from a two-element set to a two-element set, so has four possibilities. Thus there are eight arbitrary families. Naturality at \(s\) says \(\alpha_1(a)=G(s)(\alpha_0(x))=p\). The image of \(b\) is free. Exactly four families remain, indexed by the two independent choices

Naturality for identities is automatic, and \([1]\) has no other arrows, so the list is complete. None is an isomorphism: its component at \(0\) cannot be a bijection from a singleton to a two-element set. Lemma 2.1 detects the failure even when \(\alpha_1\) is bijective. \(\square\)

Exercise 2 (intermediate: what postcomposition can detect). For a nonempty category \(I\), prove that \(P\) is faithful if and only if \(P_*:\operatorname{Fun}(I,\mathcal D)\to\operatorname{Fun}(I,\mathcal E)\) is faithful. Prove the analogous equivalence for full faithfulness. Explain the exception for empty \(I\), and give a faithful functor for which nonempty postcomposition is not full.

Solution. The forward implications are Theorem 2.2. For the converse, define the constant functor \(\Delta_A\) at an object \(A\) by sending every arrow to \(1_A\). It preserves identities and composition, including those of any shape \(I\). Every arrow \(a:A\to B\) gives a constant-component natural transformation \(\Delta_a:\Delta_A\Rightarrow\Delta_B\): its naturality equation is \(1_Ba=a1_A\).

If \(P(a)=P(b)\), then \(P_*\Delta_a=P_*\Delta_b\). Faithfulness of \(P_*\) gives equality of their components. Choose one object \(i\) of nonempty \(I\); at \(i\) this is \(a=b\). Hence \(P\) is faithful. If \(P_*\) is also full and \(c:P(A)\to P(B)\), the constant transformation \(\Delta_c\) lifts to a transformation \(\alpha:\Delta_A\Rightarrow\Delta_B\). Its component at \(i\) is an arrow with \(P(\alpha_i)=c\), proving fullness of \(P\). Faithfulness has already been proved, so \(P\) is fully faithful. No connectedness assumption on \(I\) was needed.

With empty \(I\), each functor category is the one-object one-arrow category, even if its target category is empty. Postcomposition is then fully faithful for every \(P\); it detects nothing about \(P\).

For the requested example, take \(\mathcal D=[1]\) and \(\mathcal E=\mathsf{Set}\). Send \(0\) to \(\{a\}\), \(1\) to \(\{b\}\), and \(s\) to the unique map between them. This is a functor. Each Hom set of \([1]\) is empty or a singleton, so every induced Hom map is injective: the functor is faithful. It is not full, since the unique map \(\{b\}\to\{a\}\) has no preimage in \(\operatorname{Hom}_{[1]}(1,0)\). For any nonempty \(I\), that map gives a constant transformation \(P\Delta_1\Rightarrow P\Delta_0\), whereas a transformation \(\Delta_1\Rightarrow\Delta_0\) would need an impossible component \(1\to0\) at any object of \(I\). Therefore this \(P_*\) is not full. \(\square\)

Exercise 3 (advanced: a noncommutative object with a commutative center). Let \(M\) be any monoid, and let \(BM\) be its one-object category, with composition given by multiplication in \(M\). Identify \(Z(BM)\) and \(\operatorname{Aut}(\operatorname{id}_{BM})\). Apply the answer to the permutation group \(S_3\), and explain why \(\operatorname{End}_{BM}(*)\) need not be commutative.

Solution. A transformation of the identity has one component, an element \(z\in M\). Its naturality equation at \(m\in M\) is \(mz=zm\). Thus

Composition is the inherited monoid multiplication. Two elements of this set commute because either one commutes with every element of \(M\). A natural automorphism is exactly a central element invertible in \(M\), by Lemma 2.1. Its inverse is central: from \(mz=zm\), multiply by \(z^{-1}\) on both appropriate sides to get \(z^{-1}m=mz^{-1}\). Hence these elements are precisely the units of the center, as a group.

For \(M=S_3\), a central permutation \(z\) must preserve the two sets \(\{1,2\}\) and \(\{2,3\}\), since conjugating \((12)\) and \((23)\) gives \((z(1)z(2))\) and \((z(2)z(3))\). Their intersection forces \(z(2)=2\); then their remaining elements force \(z(1)=1\) and \(z(3)=3\). Thus the center and the natural-automorphism group are both trivial. Nevertheless \(\operatorname{End}_{BS_3}(*)=S_3\) is noncommutative: with rightmost permutation applied first, \((12)(23)\) sends \(1\) to \(2\), whereas \((23)(12)\) sends \(1\) to \(3\). The center imposes compatibility with every arrow; an arbitrary endomorphism of the object does not. \(\square\)

Exercise 4 (expert: diagrams with two inputs). For arbitrary categories \(A,B,D\), construct an isomorphism of categories

Give both constructions on functors and on every natural transformation. Verify the mixed-arrow composition law, and include both possible empty inputs.

Solution. The product category has objects \((a,b)\), arrows \((f,g):(a,b)\to(a',b')\), identity pairs and componentwise composition. Associativity and both identity laws hold because they hold in each coordinate. These data make sense as ambient sets by the set and product conventions of Universes and small categories.

For \(H:A\times B\to D\), define \(K(a)(b)=H(a,b)\) and \(K(a)(g)=H(1_a,g)\). The identities and composites in the second coordinate make \(K(a)\) a functor. Set

In the product, \((1_{a'},g)(f,1_b)=(f,g)=(f,1_{b'})(1_a,g)\). Applying \(H\) gives naturality of (6.5) as a transformation \(K(a)\Rightarrow K(a')\). The equations for \((1_a,1_b)\) and \((f'f,1_b)\) show that \(K\) preserves identity transformations and vertical composites, so \(K:A\to\operatorname{Fun}(B,D)\) is a functor.

Conversely, given \(K\), put \(H(a,b)=K(a)(b)\), and for a pair of arrows define

The two expressions agree by naturality of \(K(f)\). They have source \(K(a)(b)\) and target \(K(a')(b')\). Identity pairs act as identities. For a second pair \((f',g'):(a',b')\to(a'',b'')\), use naturality of \(K(f')\) at \(g\) in the middle of the composite:

Thus \(H\) is a functor. Starting from \(H\), formula (6.6) recovers it because every \((f,g)\) has the displayed product factorization. Starting from \(K\), (6.5) recovers its components since \(K(a')(1_b)\) is an identity. Its object values and arrows within \(B\) are recovered as well. So the object constructions are inverse as equalities, without choices.

For a transformation \(\tau:H\Rightarrow H'\), define the component at \(a\) to be the transformation with components \(\tau_{(a,b)}\). Naturality at \((1_a,g)\) proves it is natural in \(b\), and naturality at \((f,1_b)\) proves naturality in \(a\). Conversely, a transformation \(\sigma:K\Rightarrow K'\) has components \(\sigma_{a,b}\). These satisfy naturality for each separate type of product arrow. For a mixed arrow the full equation is

The middle equalities use the outer and inner naturality of \(\sigma\), respectively. These constructions on arrows are inverse component by component. They preserve identities and vertical composites for the same reason. This proves an isomorphism of categories in (6.4), including all natural transformations.

If \(A\) is empty, both sides are the one-object one-arrow category of empty diagrams. If \(B\) is empty, \(A\times B\) is empty and \(\operatorname{Fun}(B,D)\) is the one-object one-arrow category; there is exactly one functor from \(A\) to that category and exactly one transformation. Both sides again agree, including when \(D\) is empty. \(\square\)

References

• [Stacks, Categories] The Stacks Project authors, Categories, in the pinned AI-integrated Stacks edition. The horizontal two-path proof and the interchange proof are used through their exact open statements and proofs with the notation correspondence in Section 3. The matching pinned mathematical source records the same component square and interchange argument. Those linked sources retain the GNU Free Documentation License; they are not incorporated into the CC0 text of this lesson.
• [Universes and small categories] The preceding lesson, particularly Sections 4–5, for ambient category encoding and the complete natural-transformation smallness proof.
• [Relations and cancellation in categories] The preceding lesson, for complete ordinary category, faithful-functor and isomorphism conventions.
• [Ring actions and tensor–Hom in abelian categories] Section 1 of the lesson with that title, for the complete generic component argument proving center commutativity; the additive ring structure is a further construction.
• [Finite duality and failure of opposite density] Section 1 of the lesson with that title, for evaluation on arbitrary vector spaces and its separate finite-dimensional and completion properties.
• [Schapira, Categories and Homological Algebra] Pierre Schapira, An Introduction to Categories and Homological Algebra, author-hosted edition dated 1 March 2026, Definition 1.3.14 and Notation 1.3.15, for the natural-transformation and functor-category conventions. The complete horizontal-composition proofs used in Section 3 are attributed to Stacks above; the other constructions and all four solutions are given in this lesson.