Is an identity functor in Category of sets a function application? https://ncatlab.org/nlab/show/identity+functor

The identity functor on a category C is the functor idC:C→C that maps
each object and morphism of C to itself. The identity functors are the
identities for composition of functors in Cat.

https://ncatlab.org/nlab/show/function+application

A function f is defined by its association to each input value x
(belonging to some allowable domain of values) of an output value,
usually denoted f(x) or fx. The process of passing from f and x to
f(x) is called function application, and one speaks of applying f to x
to produce f(x).

https://ncatlab.org/nlab/show/Set
Is an identity functor in Category of sets a function application?
The reason I ask this is in programming such as F#, pipeline operator
https://riptutorial.com/fsharp/example/14158/pipe-forward-and-backward
 "Hello World" |> print

 value |> f

Now,
 value |> map(f)

is generally recognized as functor.
In this understanding, a simple function application
 value |> f

should be an identity functor, is this correct?
Thanks.
EDIT
(endo)Functor
value |> map(f)

identityFunctor (special case: map == identity)
value |> identity(f)

Therefore, identityFunctor is equivalent to
function application
value |> f 

in another notation,
f(value)

 A: It's basically fine* to think of the object part of a functor $C\to D$ as a function between the set of objects of $C$ and the set of objects of $D$. However, it's important to note that that's not all the information of the functor: it also involves a function from the set of morphisms of $C$ to the set of morphisms of $D$, together with some requirements on the morphism function and how the two functions relate.
*The concern is that $C$ and $D$ might have a proper class, not a set, of objects, but this makes no practical difference in this situation, and is certainly not of interest to a type-theoretic system like Haskell. It is perhaps worth noting that the object part of the identity functor of the category of sets is not a morphism in the category of sets, since the class of all sets is not itself a set.
A: Since $\mathbf{Set}$ is large (i.e., it has more objects than can be accounted for with a set), and since function application takes all values from one set to values in another set, a functor from $\mathbf{Set}$ can't technically be a function.
