# Why is the notion of a full subcategory invariant under equivalence?

nLab's page for a full subcategory gives the following "invariant definition" for a full subcategory:

A full subcategory-inclusion is a fully faithful functor.


What does it mean for this definition to be invariant under equivalences? I agree that if $A:\mathcal{X}\to\mathcal{C}$ is a fully faithful functor and $F,G:\mathcal{C}\leftrightarrow\mathcal{D}$ is an equivalence of categories then $FA$ is a fully faithful functor.

However, I think it is not necessarily true that if $B:\mathcal{Y}\to\mathcal{D}$ is a functor such that $GB$ is full and faithful, then $B$ is fully faithful. For example, let $\mathcal{C}$ be the category with 3 objects, $c_1,c_2,c_3$, and 6 non-identity arrows: $f_{12},f_{21},f_{13_1},f_{13_2},f_{23_1},f_{23_2}$, where the first number in the subscript is the domain and the second number is the codomain. Define $f_{13_1}f_{21}=f_{23_1}$ and similar for the other compositions. This is equivalent to the category $\mathcal{D}$ with 2 objects, $d_1,d_2$, and 2 non-identity arrows from $d_1$ to $d_2$. Now consider the subcategory of $\mathcal{C}$ consisting of $f_{13_1}, f_{23_2}$ (and the identity arrows). Its image under $F$ is full, but the original subcategory is not full. (I'd draw this, but I don't know how to draw a diagram with curved arrows on this site.)

So, why does this definition satisfy the "principle of equivalence"? If it only requires preservation under equivalences of categories, then why is this a good definition?

## 1 Answer

Your example is incorrect. Let us write $\mathcal{X}$ for the subcategory of $\mathcal{C}$ which you describe. You are then claiming that the inclusion functor $i:\mathcal{X}\to\mathcal{C}$ is not fully faithful, but the composition $Fi:\mathcal{C}\to\mathcal{D}$ is fully faithful. However, $Fi$ is not full. In particular, for instance, there is a map $Fi(c_1)\to Fi(c_3)$ which is not the image of any map $c_1\to c_3$ under $Fi$.

Note that it is true that a functor is fully faithful iff its composition with an equivalence is fully faithful. This is immediate from the fact that "fully faithful" means "bijection on Hom-sets", and a map is a bijection iff its composition with a bijection is a bijection.

• Which arrow $Fi(c_1)\to Fi(c_3)$ is missing? There are two such arrows; one is the image of $f_{13_1}$ and the other is the image of $f_{23_2}$. – alphacapture Sep 9 '18 at 0:26
• The one that is the image of $f_{23_2}$ is missing, since $f_{13_2}$ is not in $\mathcal{X}$. – Eric Wofsey Sep 9 '18 at 0:27
• I think I am misunderstanding the word "image"; what does it mean? In particular, is the other arrow $Fi(c_1)\to Fi(c_2)$ in the image? – alphacapture Sep 9 '18 at 0:29
• I think you are instead misunderstanding the definition of "full". A functor $A:\mathcal{X}\to\mathcal{Y}$ is full if for any objects $x$ and $y$ in $\mathcal{X}$ and any morphism $g:A(x)\to A(y)$, there exists $f:x\to y$ such that $g=A(f)$. In particular, note that $f$ must be a morphism between $x$ and $y$, not just any morphism of $\mathcal{X}$. – Eric Wofsey Sep 9 '18 at 0:36