(Revisted) Surjectivity: Examples for Compositions

Question

I'm asked to give an example where $f$ is surjective, but $g\circ f$ is not. I suspect that $f(x)=x$ and $g(x)=\frac{1}{x}$ will do the trick, namely for $f,g : \mathbb{R}\rightarrow \mathbb{R}$, right?

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Well, what about where $g$ is surjective...

Answer 1

Since $\operatorname{f}$ is surjective, the surjectivity of $\operatorname{g} \circ \operatorname{f}$ depends on $\operatorname{g}$ alone. In fact the image of $\operatorname{g} \circ \operatorname{f}$ is just the image of $\operatorname{g}$. You can choose any non-surjective function $\operatorname{g}$ and $\operatorname{g}\circ \operatorname{f}$ will fail to be surjective.

Your example $\operatorname{g}(x) = \frac{1}{x}$ works fine since $\operatorname{g}$ misses $0$, i.e. there is no $x$ for which $\frac{1}{x}=0$. Other simple examples include $\operatorname{g}(x) = x^2$, $\operatorname{g}(x) = \operatorname{e}^x$ or even the constant function $\operatorname{g}(x) = 0$.

Answer 2

Arguably the simplest example (as mentioned in the comments) is to take any surjective $f:\mathbb{R}\to \mathbb{R}$ and let $g:\mathbb{R} \to \mathbb{R}$ be constant.