Why does the use of functions defined in outer scope not break referential transparency?

I am learning Haskell. If I understand correctly, a simple function in Haskell is always referentially transparent. I thought it means its output depends only on the arguments passed to it.

But a function f can call another function g, defined in the outer scope. So in that sense, f's return value depends on the definition of g. And the function g is not passed to f as a parameter - at least not explicitly. Doesn't this break referential transparency?

Solution

Referential transparency means that

f s = "Hello " ++ g s
g s = s ++ "!"

is indistinguishable from

f s = "Hello " ++ h s
  where h s = s ++ "!"
g s = s ++ "!"

and

f s = "Hello " ++ s ++ "!"
g s = s ++ "!"

It means you can inline g into f without it altering the meaning of f.

If it was going to alter the meaning of f, you would have to alter g somehow. How?