Definition of functors

A functor $F: \mathcal C \to \mathcal D$ is a

• function $F : \operatorname{Ob} \mathcal C \to \operatorname{Ob} \mathcal D$ on objects and
• for each $X, Y \in \mathcal C$ a function $$F_{X,Y} : \operatorname{Hom}_{\mathcal C}(X,Y) \to \operatorname{Hom}_{\mathcal D}(FX,FY)$$

which we will often also just write as $F$ using the context to disambiguiate.

These are required to satisfy $$F( f \circ g ) = F(f) \circ F(g)$$ for any pair of morphisms $f,g$ and $$F(1_X) = 1_{FX}$$

Examples

• For any category $\mathcal C$, there is the identity function $\operatorname{Id} : \mathcal C \to \mathcal C$ which is the identity functor on objects and morphisms.

• Given a homomorphism $f : R \to S$ of rings, there is a pushforward (or restriction of scalars) functor $$f_\ast : \operatorname{Mod} S \to \operatorname{Mod} R$$ which on the underlying abelian group is $f_\ast N := N$ with the $R$-action given by $$r \cdot n := f(r) n.$$ For a morphism of $\phi : N \to N^\prime$ in $\operatorname{Mod} S$, we set $$f_\ast \phi := \phi$$ Let’s quickly see that we have a module homomorphism $$f_\ast \phi( r \cdot n ) = \phi(f(r) n) = f(r) \phi(n) = r \cdot f_\ast \phi(n)$$ where we can extract $f(r)$ since $\phi$ is assumed to be a $S$-module homomorphism.

  Thus $f_\ast$ is well-defined and it is immediate that it preserves composition and identity maps.

• Another important functor starts from a $R$-$S$ bimodule $P$. I claim that $$\operatorname{Hom}_R(P,-) : \operatorname{Mod} R \to \operatorname{Mod} S$$ is a well-defined functor.

  First we need to give the set $\operatorname{Hom}_R(P,-)$ the structure a $S$-module. As an abelian group, we set $$(f+g)(p) := f(p) + f(p)$$
  and for the $S$-module structure we let $$s \cdot f(p) := f(ps).$$

  For morphisms, we take $$\operatorname{Hom}_R(P,f)(\phi) := f \circ \phi.$$

As the Hom-functor $\operatorname{Hom}_R(P,-)$ will be quite central to our investigation, let’s work through a couple sub-examples.

• Take $R = S = P$. This bimodule is usually called the diagonal of $R$ and denoted by $\Delta_R$. We already saw that evaluation at $1$ is a bijection $$\operatorname{Hom}_R(\Delta,M) \overset{ev_1}{\to} M \\ \phi \mapsto \phi(1)$$ for any $R$-module $M$. This is in fact an isomorphism of $R$-modules since $$(r \cdot \phi)(1) = \phi(r) = r \phi(1).$$ Thus, for each $M$, application of $\operatorname{Hom}_R(\Delta, M)$ is “naturally” isomorphic to $M$. We will make this precise in a moment.

• Take $R = S = k[x]$ and $P = k[y,z]$ with the bimodule structure given by $x \mapsto y$ and $x \mapsto z$. To compute $\operatorname{Hom}_{k[x]}(k[y,z],-)$ we will write a presentation for $k[y,z]$ as a $k[x]$-module.

  As $k[x]$-module, $k[y,z]$ has a infinite basis given by $z^i$ for $0 \leq i < \infty$. Since it is a free module, $$\operatorname{Hom}_{k[x]}(k[y,z],-) \cong \operatorname{Hom}_k(k[z],-).$$ We can specify the $k[x]$-module homomorphism by providing a $k$-linear map from $k[z]$. Let’s specialize to $M = k[x]/(x) = k$. Then any linear map $k[z] \to k$ can be written as an infinite sum $$\sum_{i = 0}^\infty a_j \delta_j$$ where $$\delta_j (z^i) = \begin{cases} 1 & j = i \\ 0 & \text{ else } \end{cases}$$ The $k[x]$ module structure is given by $x \delta_i = \delta_{i-1}$ for $i > 0$ and $x \delta_0 = 0$.

Natural transformations

Given two functors $F,G : \mathcal C \to \mathcal D$, a natural transformation $\nu : F \to G$ is a collection of morphisms $$\nu_X : FX \to GX$$ for each object $X \in \mathcal C$. For any $f: X \to Y$ in $\mathcal C$, we require the diagram

commutes.

Examples

• If $F = G$, then we can take $\nu_X = 1_{FX}$. This is the identity natural transformation.
• If $F = G = \operatorname{Id}$, then saying that $\nu: F \to G$ is a natural transformation is requiring that

is requiring that $\nu_X \circ f = f \circ \nu_Y$ for any morphism $f: X \to Y$. In particular, if $f : X \to X$, then $\nu_X$ must be in the center of $\operatorname{End}(X)$.

• We have the free abelian group functor $$\mathbb{Z}^\bullet : \operatorname{Set} \to \operatorname{Ab}$$ and the forgetful functor $$\operatorname{forget}: \operatorname{Ab} \to \operatorname{Set} \\ A \mapsto A.$$ There is a natural transformation $$\operatorname{Id} \to \operatorname{forget}(\mathbb{Z}^\bullet)$$ given by $$\nu_U : U \to \mathbb{Z}^U \\ u \mapsto e_u$$ Indeed, since $$\mathbb{Z}^f\left(\sum a_u e_u\right) := \sum a_u e_{f(u)}$$ we see that for $f: U \to V$ $$\mathbb{Z}^f \circ \nu_U = e_{f(u)} = \nu_V \circ f.$$
• Given an $R$-$S$ bimodule morphism $\psi : P \to Q$, we have a natural transformation $$\operatorname{Hom}_R(Q,-) \to \operatorname{Hom}_R(P,-) \\ \phi \mapsto \phi \circ \psi$$

Functor categories

Using natural transformations, we can put functors into a category. (Remember the Caution). The objects of $\operatorname{Func}(\mathcal C, \mathcal D)$ are functors $F : \mathcal C \to \mathcal D$ while $$\operatorname{Hom}_{\operatorname{Func}(\mathcal C, \mathcal D)}(F,G) := \lbrace \nu : F \to G \mid \nu \text{ a natural transformation } \rbrace$$ Composition of natural transformation is given object-wise $$(\nu \circ \mu)_X = \nu_X \circ \mu_X$$ and the identity is the identity natural transformation.

Each object of $\mathcal C$ gives two functors. $$h_X : \mathcal C \to \operatorname{Set} \\ Y \mapsto \operatorname{Hom}_{\mathcal C}(X,Y)$$ and $$h^X : \mathcal C^{op} \to \operatorname{Set} \\ Y \mapsto \operatorname{Hom}_{\mathcal C}(Y,X)$$

A map $f : X \to X^\prime$ gives a natural transformation $$f_Y : \operatorname{Hom}_{\mathcal C}(X^\prime,Y) \to \operatorname{Hom}_{\mathcal C}(X,Y) \\ \phi \mapsto \phi \circ f.$$ and similarly $f_Y : h^X \to h^{X^\prime}$.

Taking everything together, we have two functors $$h_\bullet : \mathcal C^{op} \to \operatorname{Func}(\mathcal C, \operatorname{Set})$$ and $$h^\bullet : \mathcal C \to \operatorname{Func}(\mathcal C^{op}, \operatorname{Set}).$$

Faithful, full, essentially surjective, equivalences

• A functor $F$ is faithful if it is injective on morphism sets $$F(f) = F(g) \Rightarrow f = g$$
• $F$ is full if it surjective on morphism sets $$g = F(f)$$ for some $f$.

One can of course make the analogous definition on the level of objects but it turns out that having a natural flexibility is useful in practice.

Two objects $X,Y \in \mathcal C$ are isomorphic if there exists morphisms $f : X \to Y$ and $g : Y \to X$ such that $gf = 1_X$ and $fg = 1_Y$.

Since being isomorphic is an equivalence relation, we form the quotient $\operatorname{Ob} \mathcal C/ \sim_{iso}$ by it.

A functor $F$ is essentially surjective if the induced map $$F : \operatorname{Ob} \mathcal C \to \operatorname{Ob} \mathcal D/ \sim_{iso}$$ is surjective. In other words, every object of $\mathcal D$ is isomorphic to an object in the image of $F$.

A functor $F : \mathcal C \to \mathcal D$ is an equivalence there is a functor $G : \mathcal D \to \mathcal C$ and natural isomorphisms $$F \circ G \cong \operatorname{Id}_{\mathcal D} \\ G \circ F \cong \operatorname{Id}_{\mathcal C}$$ We will often call $G$ the inverse functor and denote it by $F^{-1}$. Note this is misleading. In general, $F^{-1}$ is only determined up to natural isomorphism.

All these definitions are tied closely together by the following fundamental result.

Theorem. $F$ is an equivalence if and only if it is full, faithful, and essentially surjective.

Proof. Let’s first assume that $F$ is an equivalence. Let $\nu : F^{-1} F \to \operatorname{Id}$ denote the natural isomorphism. We have the following diagram

and we want to check that $\varphi$ is actually the identity function. Note that, since $\nu_X$ and $\nu_Y$ are isomorphisms themselves, the conjugation $\nu_Y \circ - \circ \nu_X^{-1}$ is a bijection. Since $\nu$ is natural transformation, we know that

commutes. Thus, $$\varphi(f) = \nu_Y \circ F^{-1}F(f) \circ \nu^{-1}_X = f$$ So $F$ is injective and $F^{-1}$ is surjective. Reversing the roles of $F$ and $F^{-1}$, we see that both are bijective.

Since $X \cong FF^{-1}X$, $X$ is isomorphic to $F$ applied to $F^{-1}X$. Hence, $F$ is essentially surjective.

Now assume that $F$ is full, faithful, and essentially surjective. For each $Y \in \mathcal D$, pick some $X \in \mathcal C$ and isomorphism $\phi_Y : FX \to Y$.

Define $G : \mathcal D \to \mathcal C$ by $$GY := X$$ and $$\operatorname{Hom}_{\mathcal D}(Y,Y^\prime) \to \operatorname{Hom}_{\mathcal C}(X,X^\prime) \\ \psi \mapsto F_{X,X^\prime}^{-1} \left( \phi_{Y^\prime} \circ \psi \circ \phi_Y^{-1} \right)$$ We leave it as an exercise to check that $G$ is

• well-defined and
• an inverse to $F$.