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I've already seen an example, here, that fit this description, but I just need clarification about whether or not this is correct with these domains/codomains.

$f: \{0\} → \{0, 1\}$

$g: \{0, 1\} → \{0\}$

$g ◦ f: \{0\} → \{0\}$

$f(0) = 0$, so the codomain $\{0, 1\}$ is not equal to the range $\{0\}$, meaning $f$ is not surjective and therefore is not bijective.

$g(0) = 0$ and $g(1) = 0$, so the function is not injective, meaning $g$ is therefore not bijective.

$g(f(0)) = g(0) = 0$, so the codomain $\{0\}$ is equal to the range $\{0\}$ and the function is one-to-one, meaning $g ◦ f$ is bijective.

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  • $\begingroup$ Everything is fine. $\endgroup$ – user251257 Mar 27 '16 at 3:26
  • $\begingroup$ Very good sir. Keep it up. $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 '16 at 3:30
  • $\begingroup$ Yes. The domain of $g f$, for any functions$ f,g,$ is Dom$(g)\cap \{f(x): x\in $Dom$(f)\}.$ $\endgroup$ – DanielWainfleet Mar 27 '16 at 4:47
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Yes, you are correct. Well done.

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