Let $$f: \mathbb{R} \to \mathbb{R}$$ and $$g: \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = \begin{cases} \log_e x, & x \gt 0 \\ e^{-x}, & x \leq 0 \end{cases}$$ and $$g(x) = \begin{cases} x, & x \geq 0 \\ e^x, & x \lt 0 \end{cases}$$. Then, $$g \circ f: \mathbb{R} \to \mathbb{R}$$ is:

 $$h(x) = g(f(x))$$.

If $$x > 0$$, $$f(x) = \log_e x$$.

 If $$\log_e x \ge 0$$ (i.e., $$x \ge 1$$), $$h(x) = \log_e x$$.

If $$\log_e x < 0$$ (i.e., $$0 < x < 1$$), $$h(x) = e^{\log_e x} = x$$.

If $$x \le 0$$, $$f(x) = e^{-x} \ge 1$$.

Since $$f(x) \ge 1$$, $$h(x) = e^{-x}$$ (using the $$x \ge 0$$ case of $$g(x)$$).

The function values are always $$>0$$. Therefore, it can never reach negative values in the codomain $$\mathbb{R}$$. It is not onto.

Check values: $$h(1/2) = 1/2$$. Also, $$h(x) = e^{-x}$$ for $$x \le 0$$ covers $$[1, \infty)$$. The function is not monotonic and repeats values.

Result: B (neither one-one nor onto).