Is this true? If $g$ composed with $f$ maps $A$ onto $C$, then $f$ maps $A$ onto $B$ and $g$ maps $B$ onto $C$. I'm not sure about a counterexample in which $g$ maps $B$ onto $C$, but $f$ doesn't map $A$ onto $B$.
2026-03-31 12:52:18.1774961538
Composition of functions
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There are many counterexamples. Take any map that satisfies your rules. Add an element $x$ to set $B$ and redefine $g$ that sends $x$ to set $C$. Then, $f$ is not onto.