What does the following notation means in the context of 2-categories? $\bullet$ I know the elementary definition of an adjunction and the compact version (i.e. in terms of naturality of two hom-sets) but I am reading a definition which I cant understand its notation, namely the $\bullet$ operation of 2-arrows.
Let $F: C\rightarrow D$ and $G:D \rightarrow C$ be functors and $\eta :id_C\Rightarrow GF$and $\epsilon:FG\Rightarrow id_D$ be respectively the unite and counit of the adjunction $F\dashv
 G$. Then the following characterize this adjunction:
$$
\epsilon F \bullet F \eta =id_F \mbox{      and    } G\epsilon \bullet \eta G = id_G
$$
I would be very thankful if you let me know what does this operation means in general for 2-categories, as the notes that I am reading is using it constantly.
 A: If this notation is anything like Mac Lane's, the black dot is ordinary "vertical" composition as opposed to the "horizontal" composition denoted with a hollow circle. So those two identities are normal compositions in $D$ and $C$ respectively.
In comparison, given two functors $S,T:A\to B$ and two $S',T':B\to C$ and natural transformations $\sigma:S\to T$ and $\tau:S'\to T'$, $(\tau \circ \sigma)_c := T'\sigma_c \circ \tau S_c = \tau T_c \circ S'\sigma_c$.
A: In the notation I'm familiar with, $\cdot$ is used for the product within a (1-)category -- in particular, the hom-categories $\hom(C,D)$ (here, it is usually called "vertical composition") -- and either $\circ$ or ordinary juxtaposition is used for the product functor in a 2-category, which is usually called "horizontal composition".
Here is your arithmetic expression written as a two-dimensional formula (rather than the usual one-dimensional formulas we use in arithmetic):
$$\begin{matrix}
C &-&-&\xrightarrow{\mathbf{1}_C} & C  & \xrightarrow{F} & D
\\|| & & \Downarrow \eta& & || & \mathbf{1}_F& ||
\\C &\xrightarrow{F} &D &\xrightarrow{G} &C &\xrightarrow{F} & D
\\ || &\mathbf{1}_F& || & & \Downarrow \epsilon & & ||
\\ C &\xrightarrow{F} & D &-&-&\xrightarrow{\mathbf{1}_D} & D
\end{matrix}$$
The smaller rectangles multiply together to give the bigger rectangle, which is a 2-arrow from $F$ to $F$. Your equation claims that this big triangle is equal to $\mathbf{1}_F$.
(drawing diagrams with rectangles is, IMO, usually more convenient than the usual style)
