Find $f$ and $g$ such that domain $(f\circ g)=\mathbb{R}$ and domain $(g\circ f)=\emptyset$

Question

Find $f$ and $g$ such that domain $(f\circ g)=\mathbb{R}$ and domain $(g\circ f)=\emptyset$

That's it, I can't think of any.

I've thought of $f(x)=-1$ and $g(x)=\sqrt{x}$, and then: $$f\big(g(x)\big)=-1$$ $$g\big(f(x)\big)=\sqrt{-1}$$

Which would in principle satisfy it, but the thing is, in $f\circ g$, can I say that the domain is $\mathbb{R}$? Or is it the same as the domain of $g$?

Answer 1

That's indeed impossible.

Assuming you are talking about real functions defined on subsets of the real line.

Suppose the domain of $f\circ g$ is $\mathbb R$. Then $g$ is defined for all real numbers, and $f$ is defined at least for $x_0 = g(17) \in \mathbb R$. Now since $g$ is defined for all real numbers, it is defined at $f(x_0)$, so $g \circ f$ is defined at $x_0$, hence its domain cannot be $\emptyset$.

Answer 2

You already have a correct answer, from user334639, under the assumption that all the inputs and outputs of your functions are real numbers. On the other hand, if you're allowed to use some entity $Q$ that isn't a real number, then you can obtain an example. Let $g$ be the identity function on $\mathbb R$ and let $f$ be the constant function with domain $\mathbb R$ that maps all real numbers to $Q$. Then, $f(g(x))$ is defined (and equal to $Q$) whenever $x\in\mathbb R$, but $g(f(x))$ is never defined, because the only output of $f$ is $Q$, which is not in the domain of $g$.