Let $f\colon A\to B$, and $g\colon B\to C$.
Prove: If $g\circ f$ is onto and $g$ is 1-1, then $f$ is onto.
Here is all I can think of: $g\circ f\colon A\to C$, and for all $x\in B$ there is a unique $y\in C$.
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$\begingroup$Given an element of $B$ we need to find an element in $A$ which maps to it under $f$. Given $b\in B$, there is an element $c\in C$ such that $g(b)=c$. Since $g\circ f$ is onto, there exists an element $a\in A$ such that $g(f(a))=c$. Since $g$ is $1-1$ $f(a)=b$.
$\endgroup$ 1 $\begingroup$Since $g\circ f$ is onto, then for every $c\in C$ there exists (at least one) $a\in A$ such that $g\circ f(a) = c$.
In particular, since $f(a)=b\in B$, and $g(b)=c$, then $g$ is onto.
So $g$ is onto, and $g$ is also 1-1.
So $g$ is ......
Once you fill in the blank, use that to get that $f$ is onto.
Added: This is not the most direct way of doing it; Timothy Wagner's direct argument is more streamlined. I did it this way, because through long use, "if a $g\circ f$ is onto, then $g$ is onto" and "if $g\circ f$ is 1-1 then $f$ is 1-1" are so ingrained that they immediately jump out, just like if I see "$2\times 7$" I immediately think "$14$".
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