Let $$f, g : N \to N$$ such that $$f(n + 1) = f(n) + f(1)$$ $$\forall n \in N$$ and $$g$$ be any arbitrary function. Which of the following statements is NOT true?

Given $$f(n + 1) = f(n) + f(1)$$ for all $$n \in \mathbb{N}$$, we can deduce that $$f(n) = n \cdot f(1)$$ for all $$n \in \mathbb{N}$$ (by induction). Let $$f(1) = k$$ where $$k \in \mathbb{N}$$, so $$f(n) = kn$$.

Checking Option A: If $$f$$ is onto, then every natural number must be in the range. Since $$f(n) = kn$$, the range is $$\{k, 2k, 3k, \ldots\}$$. For this to equal $$\mathbb{N}$$, we need $$k = 1$$, so $$f(n) = n$$. This statement is TRUE.

Checking Option C: $$f$$ is one-one. Since $$f(n) = kn$$ and $$k \geq 1$$, if $$f(m) = f(n)$$ then $$km = kn$$ so $$m = n$$. This statement is TRUE.

Checking Option D: If $$f \circ g$$ is one-one, then $$g$$ is one-one. If $$g(m) = g(n)$$, then $$f(g(m)) = f(g(n))$$, and since $$f \circ g$$ is one-one, we must have $$m = n$$. This statement is TRUE.

Checking Option B: If $$g$$ is onto, then $$f \circ g$$ is one-one. Consider the counterexample where $$k = 1$$ (so $$f(n) = n$$) and define $$g$$ as: $$g(1) = g(2) = 1$$ and $$g(n) = n - 1$$ for $$n \geq 3$$. Then $$g$$ is onto (every natural number is in the range), but $$f(g(1)) = f(1) = 1 = f(1) = f(g(2))$$ while $$1 \neq 2$$, so $$f \circ g$$ is NOT one-one. This statement is NOT TRUE.

Therefore, the statement that is NOT true is: if $$g$$ is onto, then $$f \circ g$$ is one-one.