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quotient category
Quotient categories are a generalization of quotient structures found in many mathematical systems, such as sets and groups. For example, if we have a set $S=\{a,b,c\}$, by identifying $a$ and $c$ as being the “same”, we obtain a twoelement set $\{\{a,c\},b\}$, called the quotient set of $A$ by identifying $a$ and $b$. Here is a more useful example: take the set $S$ of integers. Identify integers $m$ and $n$ if their difference is divisible by $2$. Then we obtain a quotient of $S$ consisting of two elements: the set of odd numbers and the set of even numbers.
In both of the examples above, we start with a set $S$ and a binary relation $R$ (in the first example, $R$ is the singleton $\{(a,b)\}$, and in the second, $R$ consists of pairs of integers $(m,m\!+\!2n)$, where $m,n$ are integers) on $S$. The quotient $Q$ of $S$ by $R$ then satisfies the following functional properties:
1. 2. if we have another onto map $q:S\to T$ satisfying 1 above, then there is a unique map $f:Q\to T$ such that $f\circ p=q$.
The two properties above are exactly what we need to define the quotient of a category $\mathcal{C}$ by a binary relation $R$ on $\mathcal{C}$. But what is $R$ precisely? We define a binary relation on a category $\mathcal{C}$ to be a binary relation $R$ on $\operatorname{Mor}(\mathcal{C})$, the class of morphisms of $\mathcal{C}$. In other words, $R$ is a class consisting of pairs $(f,g)$ where $f$ and $g$ are morphisms in $\mathcal{C}$. If $(f,g)$ is in $R$, we write $fRg$.
Definition. Let $\mathcal{C}$ be a category and $R$ a binary relation on $\mathcal{C}$. Then a quotient of $\mathcal{C}$ by $R$ is a pair $(\mathcal{Q},P)$ where $\mathcal{Q}$ is a category and $P:\mathcal{C}\to\mathcal{Q}$ is a functor, satisfying the following two conditions:
1. $P$ is a surjection on objects; that is, for every object $Q$ in $\mathcal{Q}$, there is a object $C$ in $\mathcal{C}$, such that $P(C)=Q$.
2. if $fRg$, then $P(f)=P(g)$,
3. if $(\mathcal{D},F)$ is another pair satisfying conditions 1 and 2 above, then there is a unique functor $G:\mathcal{Q}\to\mathcal{D}$ such that $G\circ P=F$.
It is easy to see that $\mathcal{Q}$ is unique up to natural isomorphism, if it exists. The category $\mathcal{Q}$ is called the quotient category of $\mathcal{C}$ by $R$, written $\mathcal{C}/R$.
To see how $\mathcal{C}/R$ may possibly be constructed, we make the following few observations:

First, if we take the reflexive, symmetric, and transitive closure $R^{*}$ of $R$ (so that $R^{*}$ is the smallest equivalence relation containing $R$), then $fR^{*}g$ implies $P(f)=P(g)$.

In addition, if $f_{1}R^{*}g_{1}$ and $f_{2}R^{*}g_{2}$ such that $f_{1}\circ f_{2}$ and $g_{1}\circ g_{2}$ are both defined, then $P(f_{1}\circ f_{2})=P(g_{1}\circ g_{2})$. This means that we may reduce $R^{*}$ to a congruence relation $R^{{\prime}}$ with respect to morphism composition. Once again, $fR^{{\prime}}g$ implies $P(f)=P(g)$.

Furthermore, if $f:A\to B$ and $g:C\to D$ and $fR^{{\prime}}g$, then $P(f)=P(g)$ forces $P(A)=P(C)$ and $P(B)=P(D)$. In other words, the domains of $f$ and $g$ are identified under $P$. The same is true for the corresponding codomains. This implies that, if $[f]$ denotes the equivalence class of $f$ under $R^{{\prime}}$, then the domains (codomains) of all $g\in[f]$ are identified as one object in $\mathcal{C}/R$, if it exists.

As a result, all the corresponding identity morphisms are identified as well, for if $P(A)=P(B)$, then $P(1_{A})=1_{{P(A)}}=1_{{P(B)}}=P(1_{B})$. This means we can further reduce $R^{{\prime}}$ to $R^{{\prime\prime}}$ by such identifications. Note that $R^{{\prime\prime}}$ is still a congruence relation.
Note, however, the construction above does not guarantee that $\mathcal{C}/R$ in fact exists. This is mainly due to the identification of objects in $\mathcal{C}$, which may result in blowing up some hom sets into proper classes. However, if
1. $R$ is a set, or
2. if $fRg$ implies that $f$ and $g$ are parallel (they have the same domain and codomain),
then $\mathcal{C}/R$ exists. In the first case, every equivalence class with respect to $R^{{\prime\prime}}$ is a set, so that hom sets remain sets. In the second, $R^{{\prime\prime}}$ is not even needed, so that no two objects are identified, and there is no danger in hom sets being blown up into classes. In fact, the second condition is how [1] defines a quotient category. Note that in this case, $P$ is a bijection on objects between $\mathcal{C}$ and $\mathcal{C}/R$.
In the following discussion, we denote objects and morphisms in $\mathcal{C}/R$ by $[A]_{R}$ and $[f]_{R}$ (images of $A$ and $f$ under $P$), or simply $[A]$ and $[f]$ when there is no confusion.
Suppose $\mathcal{C}$ is a category with a binary relation $R$ on its morphisms. Below are some examples, as well as a nonexample:
1. Suppose $\mathcal{C}$ is generated by four objects $A,B,C,D$ and three morphisms $f:A\to B$, $g:C\to D$, and $h:B\to C$. Furthermore, suppose $R=\{(f,g)\}$. Then $\mathcal{C}/R$ exists and is generated by objects $X=[A]=[C]$ and $Y=[B]=[D]$, and morphisms $\alpha=[f]:X\to Y$ and $\beta=[h]:Y\to X$. Note that whereas $\mathcal{C}$ has only a finite number of morphisms ($f,g,h$ and four identities), the quotient $\mathcal{C}/R$ has infinitely many. For example, on $X$, the morphisms are $(\beta\alpha)^{n}$, for any nonnegative integer $n$.
2. If $R$ is defined so that $fRg$ iff $f$ is parallel to $g$, then $R$ is easily seen to be a congruence relation. Then $\mathcal{C}/R$ is the category whose objects are objects of $\mathcal{C}$, such that each hom set has at most one element. As a result, if both $\hom(A,B)$ and $\hom(B,A)$ are nonempty, then $A$ is isomorphic to $B$.
3. The two examples above are artificial, here’s a concrete one: let $\mathcal{C}$ be the category of topological spaces and continuous maps, define $R$ so that $fRg$ whenever $f$ is homotopic to $g$. Then $\mathcal{C}/R$ is the category of topological spaces with homotopic classes of continuous maps.
4. Here’s an example where $\mathcal{C}/R$ does not exist. Let $\mathcal{C}$ be a category whose collection of objects is a proper class, and there is a nonidentity morphism $h_{A}$ on each object $A$. Define $R$ so that $fRg$ iff there are objects $A,B$ with $f=1_{A}$ and $g=1_{B}$. Then if $\mathcal{C}/R$ were to exist, it would have only one object, say $X$. Furthermore, $\hom(X,X)$ contains a morphism $[h_{A}]$ for every $A$ in $\mathcal{C}$. Since $[h_{A}]=[h_{B}]$ iff $A=B$, $\hom(X,X)$ is not a set. Therefore, $\mathcal{C}/R$ is not welldefined.
References
 1 S. Mac Lane. Categories for the Working Mathematician, 2nd ed. SpringerVerlag, 1997
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Functors/compositions
In the last list, point 1, it says:
$F(h\circ f)= F(h)\circ F(f) = F(h)\circ F(g) = F(h\circ g)$, as $F$ is a (contravariant) functor. However, $h\circ g$ is not defined unless $D = \operatorname{cod}(g)=\operatorname{dom}(h) = B$, which is impossible by the assumption that $B\ne D$.
I'm probably missing something obvious, but isn't $F(h)\circ F(g) = F(h\circ g)$ required for a functor only if $h \circ g$ is defined in the first place? Otherwise you could show that any functor is injective on objects: Assume $F(A_1) = B = F(A_2)$, then $id_B \circ id_B = F(id_{A_1}) \circ F(id_{A_2}) = F (id_{A_1} \circ id_{A_2})$, which is undefined unless $A_1 = A_2$.
Re: Functors/compositions
Good point! I meant to come back and work on this entry some more... adding more examples, and discussing the difference between this definition and the one given in Mac Lane. I totally forgot about it! I will get to it later today. Thank you!
Re: Functors/compositions
The entry is now in a better shape than it was a few days ago. Improvements will be ongoing, as always. But unless there are more suggestions, I am going to leave it "as is" now.
Thanks