Suppose S and T are functors with the same codomain.
Then you can construct the *comma category* <S,T>.
Let C be the common codomain of S and T,
let D be the domain of S, and let E be the domain of T.
Suppose X in D, Y in E, and f: S(X) -> T(Y) in C.
Then (X,f,Y) is an object of <S,T>.
Suppose (Z,g,W) is another object of <S,T>.
Suppose h: X -> Z in D and i: Y -> W in E.
Suppose T(i).f = g.S(h), for a commuting square in C.
Then (f,h,i,g): (X,f,Y) -> (Z,g,W) is a morphism of <S,T>.
Note that the domain of (f,h,i,g) is (dom h, f, dom i)
and the codomain of (f,h,i,g) is (codom h, g, codom i).
Suppose (g,k,l,j): (Z,g,W) -> (V,j,U) is another morphism of <S,T>.
Then the composition (g,k,l,j).(f,h,i,g) is (f,k.h,l.i,j).

Suppose D is the empty category 0.  Since 0 has no objects,
<S,T> has no objects.  In other words, <0 -> C, T> = 0.

Suppose E is the empty category 0.  Since 0 has no objects,
<S,T> has no objects.  In other words, <S, 0 -> C> = 0.

Suppose C is the empty category 0.  Then S and T may be identified with 0.
Since C has no morphisms, <0,0> has no objects.  In other words, <0,0> = 0.
We already knew this, of course, since D and E would have to be 0.

Suppose D is a singleton, a category with a unique object and morphism.
Then S may be identified with an object X of C.
Suppose Y in E and f: X -> T(Y) in C.
Then (f,Y) is essentially an object of <X,T>.
Suppose (g,W) is another object of <X,T>.
Suppose i: Y -> W in E.  Suppose T(i).f = g.
Then (f,i,g): (f,Y) -> (g,W) is a morphism of <X,T>.
Note that the domain of (f,i,g) is (f, dom i) 
and the codomain of (f,i,g) is (g, codom i).
Suppose (g,l,j): (g,W) -> (j,U) is another morphism of <X,T>.
Then (g,l,j).(f,i,g) is (f,l.i,j).
This is the usual kind of comma category.

Suppose E is a singleton, so T is identified with an object Y of C.
Suppose X in D and f: S(X) -> Y in C.
Then (X,f) is an object of <S,Y>.
Suppose (Z,g) is another object of <S,Y>.
Suppose h: X -> Z in D.  Suppose f = g.S(h).
Then (f,h,g): (X,f) -> (Z,g) is a morphism of <S,Y>.
Note that the domain of (f,h,g) is (dom h, f)
and the codomain of (f,h,g) is (codom h, g).
Suppose (g,k,j): (Z,g) -> (V,j) is another morphism of <S,Y>.
Then (g,k,j).(f,h,g) is (f,k.h,j).
This is the usual kind of cocomma category.

Suppose D and E are both singletons,
so S and T are identified with objects X and Y of C.
Suppose f: X -> Y in C.  Then f is an object of <X,Y>.
Suppose g is another object of <X,Y>.
The only morphism f -> g in <X,Y> is (f,g),
and that exists only if f = g.
In other words, <X,Y> is the set of morphisms from X to Y.

Suppose C is a singleton.
Suppose X in D and Y in E.  Then (X,Y) is an object of <S,T>.
Suppose (Z,W) is another object of <S,T>.
Suppose h: X -> Z in D and i: Y -> W in E.
Then (h,i): (X,Y) -> (Z,W) is a morphism of <S,T>.
Note that the domain of (h,i) is (dom h, dom i)
and the codomain of (h,i) is (codom h, codom i).
Suppose (k,l): (Z,W) -> (V,U) is another morphism of <S,T>.
Then (k,l).(h,i) is (k.h,l.i).
In other words, <D -> 1, E -> 1> is the Cartesian product of D and E.

Suppose S is the identity functor on C.  Then S may be identified with C.
Suppose X in C, Y in E, and f: X -> T(Y) in C.
Then (f,Y) is an object of <C,T>.
Suppose (g,W) is another object of <C,T>; let Z be dom g.
Suppose h: X -> Z in C and i: Y -> W in E.  Suppose T(i).f = g.h.
Then (f,h,i,g): (f,Y) -> (g,W) is a morphism of <C,T>.
Note that the domain of (f,h,i,g) is (f, dom i)
and the codomain of (f,h,i,g) = (g, codom i).
I suppose <C,T> could be called the slice category of C over T.

Suppose T is the identity functor on C, so T is identified with C.
Suppose X in D, Y in C, and f: S(X) -> Y in C.
Then (X,f) is an object of <S,C>.
Suppose (Z,g) is another object of <S,C>; let W be codom g.
Suppose h: X -> Z in D and i: Y -> W in C.  Suppose i.f = g.S(h).
Then (f,h,i,g): (X,f) -> (Z,g) is a morphism of <S,C>.
Note that the domain of (f,h,i,g) is (dom h, f)
and the codomain of (f,h,i,g) is (codom h, g).
I suppose <S,C> could be called the coslice category of C under S.

Suppose S and T are both the identity functor on C,
so they are both identified with C.
Suppose X and Y in C and f: X -> Y in C.
Then f is an object of <C,C>.
Suppose g is another object of <C,C>;
let Z be dom g, and let W be codom g.
Suppose h: X -> Z and i: Y -> W in C.  Suppose i.f = g.h.
Then (f,h,i,g): f -> g is a morphism of <C,C>.
Note that the domain of (f,h,i,g) is f
and the codomain of (f,h,i,g) is g.
In other words, <C,C> is the arrow category of C.

Suppose S is the identity functor on C and E is a singleton,
so S is identified with C and T is identified with an object Y in C.
Suppose X in C and f: X -> Y in C.  Then f is an object of <C,Y>.
Suppose g is another object of <C,Y>; let Z be dom g.
Suppose h: X -> Z is a morphism of C.  Suppose f = g.h.
Then (f,h,g): f -> g is a morphism of <C,Y>.
Note that the domain of (f,h,g) is f and the codomain of (f,h,g) is g.
Suppose (g,k,j): g -> j is another morphism of <C,Y>.
Then (g,k,j).(f,h,g) is (f,k.h,j).
In other words, <C,Y> is the slice category of C over Y.

Suppose D is a singleton and T is the identity functor on C,
so S is identified with an object X in C and T is identified with C.
Suppose Y is an object of C and f: X -> Y is a morphism of C.
Then f is an object of <X,C>.
Suppose g is another object of <X,C>; let W be codom g.
Suppose i: Y -> W is a morphism of C.  Suppose i.f = g.
Then (f,i,g): f -> g is a morphism of <C,X>.
Note that the domain of (f,i,g) is f and the codomain (f,i,g) is g.
Suppose (g,l,j): g -> j is another morphism of <X,C>.
Then (g,l,j).(f,i,g) is (f,l.i,j).
In other words, <X,C> is the coslice category of C under X.

Suppose C is a singleton and D = C.  Then S may be identified with 1.
Suppose Y is an object of E.  Then Y is an object of <1,T>.
Suppose W is another object of <1,T>.  Suppose i: Y -> W in E.
Then i: Y -> W is a morphism of <1,T>.
Note that the domain and codomain of i in <1,T> are the same as in E.
Suppose l: W -> U is another morphism of <1,T>.
Then l.i is the same in <1,T> as in E.
In other words, <1, E -> 1> is the category E.
We already knew this, since 1 x E = E.

Suppose C is a singleton and C = E, so T is identified with 1.
Suppose X is an object of D.  Then X is an object of <S,1>.
Suppose Z is another object of <S,1>.  Suppose h: X -> Z in D.
Then h: X -> Z is a morphism of <S,1>.
Note that the domain and codomain of h in <S,1> are the same as in D.
Suppose k: Z -> V is another morphism of <S,1>.
Then k.h is the same in <S,1> as in D.
In other words, <D -> 1, 1> is the category D.
We already knew this, since D x 1 = D.

Suppose C is a singleton and D = C = E, so S and T are identified with 1.
There is a unique object of <1,1>.  There is a unique morphism of <1,1>.
In other words, <1,1> is a singleton.
We already knew this, of course, in several ways.