let $F : \mathcal{C} \rightarrow \mathcal{D}$ be a functor from two categories. It looks like that there are various notions of kernels one could define for a functor.

One could define the arrow kernel of $F$ to be the subcategory of $\mathcal{C} \times \mathcal{C}$ with set of objects $(x,y)$ such that $F(x) = F(y)$ and set of arrows $(f, g)$ such that $F(f) = F(g)$. This is clearly a subcategory because $(Id_x, Id_y)$ is mapped to the identity, while if $(f,g)$ and $(l,m)$ are such that $F(f) = F(g)$, and $F(l) = F(m)$, then clearly $(fl, gm)$ is such that $F(fl) = F(gm)$ by functoriality. Composition is clearly associative. In particular, the arrow kernel of $F$ can be seen as a "generalized equivalence relation" on the set of objects and maps (see it as a reflexive, transitive, and symmetric generalized equivalence relation).

Indeed, this makes us want to define what a generalized equivalence relation on $\mathcal{C}$ is as a functor $R$ from $\mathcal{C} \times \mathcal{C}$ to the category with two objects and one arrow $i_0$ and $i_1$ per objects. The latter arrows play the role of $0$ for arrows. (Note that we obviously force that $i_0$ and $i_1$ are absorbing elements, that is, $i_j^2 = i_j$ for $j = 0, 1$)

We say that such a functor is :

1) reflexive on objects if for all objects $x$ in $\mathcal{C}$, $R(x,x) = 1$,

2) symmetric on objects if for all objects $x$, $y$, $R(x,y) = 1$ implies $R(y,x) = 1$,

3) transitive on objects if for all objects $x$, $y$, $z$, $R(x,y) = 1$ and $R(y,z) = 1$ implies that $R(x,z) = 1$.

1') reflexive on arrows if for all $f$ such that $dom R(f,f) = 1$ and $cod (f,f) = 1$, $R(f,f) = Id_1$. (That is, we pick up all arrows we can),

2') symmetric on arrows if for all maps $f$, $g$, $R(f,g) = Id_1$ implies that $R(g,f) = Id_1$

3') transitive on arrows if for all maps $f$, $g$, and $h$, $R(f,g) = Id_1$ and $R(g,h) = Id_1$ implies that $R(f,h) = Id_1$.

The above kernel we defined for a category is nothing else than such generalized equivalence relation $R(x,y) = 1$ iff $F(x) = F(y)$, and $R(f,g) = Id_1$ iff $F(f) = F(g)$. In particular, note that if we had a groupoid, all arrows being invertible, it would be enough to consider the full subcategory of $\mathcal{C}$ consisting of arrows $f$ such that $F(f) = Id$.

Is there a name for such generalized equivalence relations, kernels, and some applications?

PS : I didn't check yet if the above definition would allow us to speak of partition of a category in terms of equivalence classes, for relations that respects 1), 2), 3), and their arrows analogs.

edit: we put $f_1 R f_2$ to mean $R(f_1, f_2) = Id_1$ and similarly for objects. Clearly, a generalized equivalence relation that respects 1), 2), 3), and 1'), 2'), 3') is such that it partitions the objects of $\mathcal{C}$. Moreover, suppose $f_1 R f_2$ and $g_1 R g_2$. By functoriality, $f_1 R f_2$ and $g_1 R g_2$ implies $f_1 \circ g_1 R f_2 \circ g_2$ when it is defined. In particular if the relation is fully reflexive (that is 1) ), then any arrow $f$ belongs to (at least) one arrow equivalence class by 1'). Moreover, any arrow belongs only to one such. Indeed, suppose that $f$ belong to $R[g] = \{ f, fRg\}$ and $R[g']$. Then, clearly $fRg$ and $fRg'$, that is, $gRg'$, and thus, $R[g] = R[g']$.

Now, one can easily define quotient of category by a generalized equivalence relation. Indeed, let $\mathcal{C}$ be a category and $R$ a generalized equivalence relation on $\mathcal{C}$. We define the quotient category of $\mathcal{C}$ as $\mathcal{C} /R$, that is, a category whose objects are equivalence classes of objects $R[x] = \{y, yRx\}$ and arrows $R[f] = \{g, gRf\}$, and show that it is a category.

Indeed, we define a composition on the set of equivalence class as follow : $R[f] \circ R[g] = R[h]$ iff there exists two morphisms $(f_1, f_2)$ in $R[f]$, two morphisms $(g_1, g_2)$ in $R[g]$ such that $(f_1,f_2) \circ (g_1, g_2)$ is composable and equals to $(h_1, h_2)$ in $R[h]$. This clearly defines an associative composition that admits units fo each equivalence classes, hence, a category.

edit2: it would be interesting to check that the first isomorphism theorem holds for quotient category. Let $\mathcal{C}$ be a functor, and let $R$ be a generalized equivalence relation on $\mathcal{C}$. Let us call canonical projection $\Pi : F \rightarrow F/R$ the functor that sends objects and arrows on their respective equivalence class. We want to show that for any functor $F: \mathcal{C} \rightarrow \mathcal{D}$ that is constant on the equivalence classes of $R$, there exists a unique functor $\Phi$ such that $F = \Phi \circ \Pi$.

step 1: show that $\Pi$ is indeed a functor for the above composition. We clearly have $\Pi[id_x] = 1_{[x]}$, and $\Pi(h \circ f) = [g \circ f] = \{h, R(h, g \circ f) = R(h,g) \circ R(h,f)\} = [g] \circ [f]$ by functoriality.

step 2: construct the functor $\Phi$ and show that it is unique. Clearly, we send an equivalence class on one of its representative, that is $\Phi([x]) = F(x), \Phi([f]) = F(f)$ which is trivialy a functor. It is also clearly unique.

Hence, first isomorphism theorem. Is there any mistake?

We could now define the kernel of $F$ as the generated generalized (fully) reflexive equivalence relation $R$ such that R(x,y) = 1 iff $F(x) = F(y)$ and $fRg$ iff $F(f) = F(g)$, and find a first isomorphism theorem for functor.

One would maybe want to weaken our first isomorphism theorem by imposing that the functor is not constant on the equivalence classes but only constant up to isomorphism. I think that the above proof can easily be modified by adding a natural isomorphism between the functor that is indeed constant on equivalence class and such an $F$. Conjugate everything by its components and it would work.

PPS: if one wants to suppress all set theoretical references, then consider that an equivalence class is a functor $R(x,-)$ or $R(f,-)$. An equivalence relation being a (bi)functor, each components is functorial too.

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    $\begingroup$ Clearly, if $F : E \rightarrow B$ is a functor of groupoid, then the arrow kernel restricted to the objects $x$ of $E$ such that $F(x) = y$ is the fiber category over $y$. We could call it the arrow kernel of $F$ over $y$ :< $\endgroup$ – sure Mar 1 '15 at 0:18
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    $\begingroup$ There is a notion of kernel pair in any category with finite limits. More generally, we may define congruences and their quotients. You are considering the special case of the category of categories. $\endgroup$ – Zhen Lin Mar 1 '15 at 12:17
  • $\begingroup$ ah nice, I didn't know this concept. You don't need any limits to speak about my definition, isn't it convenient in this respect ? $\endgroup$ – sure Mar 1 '15 at 12:37
  • $\begingroup$ For example, I think that it is possible to make sense of equivalence relations up to isomorphisms, that is, you don't force equality, but only equality up to isomorphisms. It looks like a kernel pair can't do that. $\endgroup$ – sure Mar 1 '15 at 12:56
  • $\begingroup$ You can do a "two-dimensional" version of this in the 2-category of categories. Things become much more complicated, however. $\endgroup$ – Zhen Lin Mar 1 '15 at 14:10

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