The following question was asked in an examination:
Let $f$ and $g$ be two functions with domain and codomain equal to the set of real numbers. If,
$$g\circ f(x) = \begin{cases} x^2, & \text{if $x\geq0$} \\ e^x-1, & \text{if $x<0$} \end{cases} $$
Then choose the correct option out of the following:
(a) $f$ is one-one
(b) $f$ is onto
(c) $g$ is one-one
(d) $g$ is onto
Here, we have been asked to find whether the functions $f$ and $g$ are injective or surjective from their composite function $g\circ f$. Since, I didn't find any way out, I assumed $f(x)=x$ and $g(x)=x^2$ when $x\geq0$ and $g(x)=e^x-1$ when $x<0$. This assumption also satisfies the condition given in the question.
Clearly, $f$ is both one-one and onto. On constructing the graph of $g$ it can be seen that it's one-one but not onto. Based on this, options (a), (b) as well as (c) are correct. But the question clearly states only one of the options is correct. So, it would be helpful if someone could explain the reason for the fallacy of my argument. Also, in general, how can we determine the nature of two functions from their composite function? The only way I could think of is to decompose the composite function into component functions. But I don't think it's a good approach as it can lead to more than one set of functions.
We have constructed an example where $f$ is one-to-one, $f$ is onto, and $g$ is one-to-one.
But given another $f$ and $g$, the properties need not hold.
Consider the following functions:
Let $f(x) = \begin{cases} x^2 &, x \ge 0 \\ e^x-1, &, x<0\end{cases}$ and $g(x)=\begin{cases}x &, x \ge -1 \\ -x-2&, x < -1\end{cases}$
In that case, we see that $g$ is not one-to-one and not onto while $f$ is not onto.
By elimination, we know that $f$ is one-to-one.
Now, let's prove it.
If $$f(x)=f(y)$$ $$g\circ f(x) = g \circ f(y)$$
but $g \circ f$ is injective.
Hence $x=y$.
Hence $f$ must be one-to-one.