# The Unapologetic Mathematician

## Full and Faithful Functors

We could try to adapt the definitions of epics and monics to functors, but it turns out they’re not really the most useful. Really what we’re most concerned with in transforming one category into another is their morphisms. The more useful notions are best phrased specifically with regard to the hom-sets of the categories involved.

Remember that a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ between two (locally small) categories defines a function for each hom set: $F_{X,Y}:\hom_\mathcal{C}(X,Y)\rightarrow\hom_\mathcal{D}(F(X),F(Y))$. We now apply our standard definitions to these functions.

We say that a functor is “full” if for every pair of objects $X$ and $Y$ in $\mathcal{C}$ the function $F_{X,Y}$ is surjective. Similarly, we say that a functor is “faithful” if each function $F_{X,Y}$ is injective. When we apply these definitions to monoids (considered as categories with one object) we recover monomorphisms and epimorphisms of monoids, so we know we’re on the right track.

Now a subcategory $\mathcal{S}$ of a category $\mathcal{C}$ consists of a subclass of the objects of $\mathcal{C}$ and a subclass of the morphisms of $\mathcal{C}$ so that

• For each arrow in $\mathcal{S}$ both its source and target objects are in $\mathcal{S}$.
• For each object in $\mathcal{S}$ its identity arrow is in $\mathcal{S}$.
• For each pair of composable arrows in $\mathcal{S}$ their composite is in $\mathcal{S}$.

These conditions are enough to ensure that $\mathcal{S}$ is a category in its own right. There is an inclusion functor $I:\mathcal{S}\rightarrow\mathcal{C}$ sending each object and morphism of $\mathcal{S}$ to itself in $\mathcal{C}$. Clearly it’s faithful, since any two distinct arrows in $\mathcal{S}$ are distinct in $\mathcal{C}$.

What if this inclusion functor is also full? Then given any two objects $X$ and $Y$ in $\mathcal{S}$, every morphism $f:X\rightarrow Y$ in $\mathcal{C}$ is included in $\mathcal{S}$. We say that $\mathcal{S}$ is a “full subcategory” of $\mathcal{C}$. In practice, this means that we’ve restricted which objects of $\mathcal{C}$ we care about, but we keep all the morphisms between the objects we keep. For example, the category $\mathbf{Ab}$ of abelian groups is a full subcategory of $\mathbf{Grp}$.

Note that we have not given any conditions on the map of objects induced by a functor. In fact, the (unique) functor from the category $\mathbf{2}$ to the category $\mathbf{1}$ is faithful, even though it appears to lose a lot of information. Indeed, though we identify both objects from $\mathbf{2}$ in the image, the function on each hom-set is injective. Similarly, either functor from $\mathbf{1}$ to $\mathbf{2}$ is full. One condition on the object map that’s useful is that a functor be “essentially surjective”. A functor $F$ has this property if each object $D$ in the target category is isomorphic to an object $F(C)$ in the image of $F$.

Now, let functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ furnish an equivalence of categories. I say that $F$ (and, by symmetry, $G$) is full, faithful, and essentially surjective. Indeed, since $F\circ G\cong1_\mathcal{D}$ we have an isomorphism $F(G(D))\cong D$ for each object $D\in\mathcal{D}$, so $F$ is essentially surjective. The natural isomorphism $\eta:G\circ F\rightarrow 1$ says that for every arrow $f:C_1\rightarrow C_2$ we have $G(F(f))=\eta_{C_2}^{-1}\circ f\circ\eta_{C_1}$. The right side of this equation is a bijection from $\hom_\mathcal{C}(C_1,C_2)$ to $\hom_\mathcal{C}(G(F(C_1)),G(F(C_2)))$, and the left side is the composite $G_{F(C_1),F(C_2)}\circ F_{C_1,C_2}$. If $G$ were not full then the left side could not be always surjective, while if $F$ were not faithful then the right side could not be always injective. Since the right side is always a bijection, $G$ must be full and $F$ must be faithful. By symmetry $G$ is essentially surjective and faithful, and $F$ is full.

On the other hand, if a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ is full, faithful, and essentially surjective, then it is one side of an equivalence of categories. The proof is a bit more involved, but basically it proceeds by building a functor $G$. Since each object of $\mathcal{D}$ is isomorphic to some object in the image of $F$ we pick such an object $F(C)\cong D$ for each $D$ and set $G(D)=C$. Then we have to find mappings of the hom-sets of $\mathcal{D}$ in a coherent way so that the resulting composites $F\circ G$ and $G\circ F$ are naturally isomorphic to the identity functors. I won’t fill in all the details here, but suffice it to say it can be done.

This, then, characterizes equivalences of categories the way injectivity and surjectivity of functions characterize isomorphisms of sets.

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Posted by John Armstrong | Category theory

## 10 Comments	»

1. [...] now for the coup de grâce: the Yoneda embedding is fully faithful! Firstly, this means that the homomorphism is injective — its kernel is trivial — so [...]

Pingback by | June 8, 2007 | Reply

2. It seems to me that the exposition in the third-from-last paragraph is a bit mixed up. The isomorphism F\circ G\cong1_\mathcal{D} is what makes F essentially surjective.
\eta is reversed from what it usually is, and the positions of C_1 and C_2 need to be swapped in G(F(f))=\eta_{C_1}\circ f\circ\eta_{C_2}^{-1}; so I’d suggest reformulating this to G(F(f))=\eta_{C_2}\circ f\circ\eta_{C_1}^{-1}. Then all goes fine afaics.

Comment by Avery Andrews | September 22, 2007 | Reply

3. Avery, you’re right that I added the phrase about essential surjectivity to the wrong sentence, but I reversed the usual direction of $\eta$ (and the ensuing alterations to compensate) for a reason: this makes it clearer that an equivalence is actually a type of adjunction. I can flip the directions of these natural isomorphisms because they’re isomorphisms, but once I weaken them to mere natural transformations I no longer have that freedom.

Comment by | September 24, 2007 | Reply

4. Right, but what about \eta_{C_2}\circ f\circ\eta_{C_1}^{-1} vs \eta_{C_1}\circ f\circ\eta_{C_2}^{-1}; the latter is still looking wrong to me. Too bad categorists don’t use some sort of postfix notation for function-like things.

Comment by Avery Andrews | September 24, 2007 | Reply

5. Yes, that composition was also backwards.. A postfix notation would be useful, but unfortunately everyone else uses this ordering, and if I started out using the other then everyone reading my stuff would be confused when they try to read other material.

Comment by | September 24, 2007 | Reply

6. oops, \eta_{C_2}^{-1}\circ f\circ\eta_{C_1}, if \eta is from GF to 1_{\mathcal C}.

Comment by Avery Andrews | September 24, 2007 | Reply

7. Incidentally, Avery, try wrapping your LaTeX like this:

::dollar-sign::latex foo::dollar sign::

It comes up: $foo$. And it works on all WordPress-hosted weblogs!

Comment by | September 24, 2007 | Reply

8. cool. Maybe a `guide for commenters’ would be a useful addition?

Comment by Avery Andrews | September 24, 2007 | Reply

9. [...] show this, we must show that it is full, faithful, and essentially surjective. The first two conditions say that given [...]

Pingback by | June 24, 2008 | Reply

10. [...] , but with only one object. So we should really be thinking about the category of algebras as a full sub-2-category of the 2-category of categories enriched over [...]

Pingback by | November 18, 2008 | Reply

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