Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying the concept of faithful functors, but I cannot grasp the difference between being faithtful and being injective on arrows. Could someone explain the difference and provide some examples?

share|cite|improve this question
"Being injective on arrows" is an evil property (not invariant under equivalences of categories). "Faithful" is a good property. – Martin Brandenburg Jun 19 '14 at 10:36
up vote 5 down vote accepted

Let's remind ourselves about the definition of a faithful functor: Any functor $$ \mathcal{F} \colon \mathcal{C} \rightarrow \mathcal{D}$$ induces a map on hom-sets $$\mathcal{F}_{X,Y} \colon \mathrm{Hom}_\mathcal{C}(X,Y) \rightarrow \mathrm{Hom}_\mathcal{D}(\mathcal{F}(X),\mathcal{F}(Y))$$ for every pair of objects $X,Y$ in $\mathcal{C}$. Then $\mathcal{F}$ is called faithful if the induced maps $\mathcal{F}_{X,Y}$ are injective for all objects $X,Y$ in $\mathcal{C}$.

The crucial thing to observe is that we are quantifying over objects in the source category $\mathcal{C}$ --- this is what makes the difference between a faithful functor and one injective on morphisms. For instance consider the four-object category $\mathcal{C}$ with $\mathrm{Obj}(\mathcal{C})=\{A,B,C,D\}$ and morphisms given by $$ \mathrm{Hom}(A,B) = \{f_i\}_{i\in\mathbb{Z}}, \\ \mathrm{Hom}(C,D) = \{g_i\}_{i\in\mathbb{Z}}$$ (and for simplicity say the other hom-sets are empty, except for the identity morphisms on each object). If we have another category $\mathcal{D}$ consisting of two objects $P,Q$ and morphisms $$ \mathrm{Hom}(P,Q) = \{h_i\}_{i\in\mathbb{Z}}$$ then the functor $ \mathcal{F} \colon \mathcal{C} \rightarrow \mathcal{D}$ sending $$ A \mapsto P, \\ B \mapsto Q, \\ C \mapsto P, \\ D \mapsto Q $$ and $$ f_i \mapsto h_i, \\ g_i \mapsto h_i$$ (and identity morphisms to identity morphisms, etc.) is faithful, since restricting to every pair of objects $X,Y$ in $\mathcal{C}$ gives an injective map on $\mathrm{Hom}(X,Y)$. On the other hand $\mathcal{F}$ is neither injective on objects or morphisms.

I wanted to emphasize the fact that quantifying over objects in the source category is the important bit: if we required a faithful functor to be injective on the hom-sets between every pair of objects in the target category we would indeed just end up with a functor injective on morphisms.

share|cite|improve this answer

A functor that is injective on morphisms is automatically faithful, of course; but by thinking about identity morphisms, one sees that a functor that is injective on morphisms must also be injective on objects.

Exercise. Show that a functor is injective on morphisms if and only if it is injective on objects and faithful.

Once you have proved the claim, you will find many examples. For instance, if $\mathcal{P}$ is a poset considered as a category, then the unique functor $\mathcal{P} \to \mathbb{1}$ is always faithful, hence injective on morphisms if and only if it is injective on objects.

share|cite|improve this answer
Oh, I see what you mean. The point is that a functor can identify what were previously different sources and targets. My mistake. – Qiaochu Yuan Jun 19 '14 at 6:35
I guess the analogous result holds with injective replaced by surjective and faithful replaced by full? There any functor $F:\varnothing\to\mathcal{D}$ would serve as corresponding nonexample? – Alexander Frei Jun 28 at 8:17
Yes, that's right. – Zhen Lin Jun 28 at 9:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.