Remark 1.2.2.a in Tom Leinster's Category Theory notes 
The definition of a functor is set up so that from each string
$$ A_o \overset{f_1}{\to} ... \overset{f_n}{\to} A_n$$
of maps in $\mathcal{A}$ (with $n \geq 0$); it is possible to construct exactly one map
$$ F(A_o) \to F(A_n)$$
in $\mathcal{B}$. For example, given maps:
$$A_o \overset{f_1}{\to} A_1 \overset{f_2}{\to} A_2 \overset{f_3}{\to} A_3 \overset{f_4}{\to} A_4$$
in $\mathcal{A}$, we can construct maps:

in $\mathcal{B}$ but the axioms imply they are equal

Could someone explain what is being said here in a more simple way? I don't quite get what the the double arrow and things above and below it mean.
 A: Starting with a chain of four composable morphisms
$$
  A_0
  \xrightarrow{\; f_1 \;} A_1
  \xrightarrow{\; f_2 \;} A_2
  \xrightarrow{\; f_3 \;} A_3
  \xrightarrow{\; f_4 \;} A_4
$$
in $\mathscr{A}$, there are multiple ways of constructing an induced morphism from $F(A_0)$ to $F(A_4)$ in $\mathscr{B}$.
Leinster gives the following two explicit examples.

*

*We can start by composing the two morphisms $f_1$ and $f_2$, and also composing the two morphisms $f_3$ and $f_4$.
$$
  A_0
  \xrightarrow{\enspace f_2 f_1 \enspace} A_2
  \xrightarrow{\enspace f_4 f_3 \enspace} A_4
$$
We then apply the functor $F$ to these two composites, resulting the the two morphisms $F(f_2 f_1)$ and $F(f_4 f_3)$.
$$
  F(A_0)
  \xrightarrow{\enspace F(f_2 f_1) \enspace} F(A_2)
  \xrightarrow{\enspace F(f_4 f_3) \enspace} F(A_4)
$$
Finally, we compose the two morphisms $F(f_2 f_1)$ and $F(f_4 f_3)$.
$$
  F(A_0) \xrightarrow{\enspace F(f_4 f_3) F(f_2 f_1) \enspace} F(A_4)
$$


*We could also start by adding the identity morphisms of $A_4$ to our chain of composable morphisms.
$$
  A_0
  \xrightarrow{\; f_1 \;} A_1
  \xrightarrow{\; f_2 \;} A_2
  \xrightarrow{\; f_3 \;} A_3
  \xrightarrow{\; f_4 \;} A_4
  \xrightarrow{\enspace 1_{A_4} \enspace} A_4
$$
We then compose the two morphisms $f_2$ and $f_3$.
$$
  A_0
  \xrightarrow{\; f_1 \;} A_1
  \xrightarrow{\enspace f_3 f_2 \enspace} A_3
  \xrightarrow{\; f_4 \;} A_4
  \xrightarrow{\enspace 1_{A_4} \enspace} A_4
$$
Next, we apply the functor $F$ to all four morphisms.
$$
  F(A_0)
  \xrightarrow{\enspace F(f_1) \enspace} A_1
  \xrightarrow{\enspace F(f_3 f_2) \enspace} A_3
  \xrightarrow{\enspace F(f_4) \enspace} A_4
  \xrightarrow{\enspace F(1_{A_4}) \enspace} A_4
$$
Finally, we compose all the morphisms.
$$
  F(A_0)
  \xrightarrow{\enspace F(1_{A_4}) F(f_4) F(f_3 f_2) F(f_1) \enspace}
  F(A_4)
$$
We have now constructed two morphisms from $F(A_0)$ to $F(A_4)$, namely
$$
  F(f_4 f_3) F(f_2 f_1)
  \quad\text{and}\quad
  F(1_{A_4}) F(f_4) F(f_3 f_2) F(f_1) \,.
$$
These two morphisms have the same domain and the same codomain, and are therefore ‘parallel’.
This is why Leinster, in the last diagram, literally depicts them as two parallel arrows from $F(A_0)$ to $F(A_4)$.
However, the properties of a functor ensure that these two morphisms from $F(A_0)$ to $F(A_4)$ are actually equal.
More explicitly, we have the chain of equalities
\begin{align*}
  {}&
  F(1_{A_4}) F(f_4) F(f_3 f_2) F(f_1)
  \\
  ={}&
  1_{F(A_4)} F(f_4) F(f_3) F(f_2) F(f_1)
  \\
  ={}&
  F(f_4) F(f_3) F(f_2) F(f_1)
  \\
  ={}&
  F(f_4 f_3) F(f_2 f_1) \,.
\end{align*}
Remark 1.2.2, (a) claims that this is not just a coincidence, occurring to this specific example, but a general principle:
in whatever way we construct from a chain of morphisms
$$
  A_0
  \xrightarrow{\; f_1 \;}
  A_1
  \xrightarrow{\; f_2 \;}
  \dotso
  \xrightarrow{\; f_n \;}
  A_n
$$
in $\mathscr{A}$ a morphism
$$
  F(A_0) \longrightarrow F(A_n)
$$
in $\mathscr{B}$, the result will always be the same.
