3
$\begingroup$

Suppose $f: A \to B$ and $g: C \to D$ are functions, and that $B ⊆ C$.

I need to come up with an example where $g \circ f$ is surjective but $f$ is not. I'm confused on how exactly to do that, but I understand that to show something is surjective you have to have the range of the function equal to the codomain of the function. Here's my attempt:

So if $g(x) = x^2$ with a codomain of $ℝ$, and $f(x) = \sqrt{x}$ with a codomain of $ℝ$, then $g(f(x)) = (\sqrt{x})^2 = x$. Since the range of this function and codomain are the same, this is surjective.

But, since $f(x) = \sqrt{x}$, its range is only $[0, ∞)$, which is smaller than its codomain of $ℝ$, meaning $f$ is not surjective.

$\endgroup$
  • 2
    $\begingroup$ This is harder with $\mathbb R$ than it might be with some very simple sets. $\endgroup$ – TokenToucan Mar 19 '16 at 23:36
  • 1
    $\begingroup$ @Chris Your example is correct if you admit $A=[0,+\infty)=D$ and $B=\mathbb{R}=C$ $\endgroup$ – Darío G Mar 19 '16 at 23:39
4
$\begingroup$

Put $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=x^2$, and $g:\mathbb{R}\to \{0\}$ defined by $g(x)=0$. Then $g\circ f:\mathbb{R}\to \{0\}$ is clearly surjective but $f$ is not surjective.

$\endgroup$
3
$\begingroup$

Let $B=C=\{1,2\}$, and $A=D=\{3\}$. Then pick any functions $f,g$ you want, to get $g\circ f$ surjective but $f$ not surjective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.