Let $f:A→B$ and $g:B→C$ be maps. Prove that if $g∘f$ is one to one and f is surjection, then g is one to one.
we want to proof that $ g(b) = g(b')$ for $b,b'\in B$ ,$b=b'$
since f is onto $ f(a)=b$ for $ b \in B $ and $f(a')=b'$ for $ b'\in B$
since $gof$ is injection, there exist $a,a' \in A$ , $gof(a)=gof(a')$ such that $a=a'$
$g(f(a))=g(f(a'))$
$g(b)=g(b')$
$b=b'$
hence there exist $b,b' \in B$ such that $g(b)=g(b')$ as required
is my proof correct?
if there is a better way i would like to know thanks
Best Answer
Your proof is correct...ish: that $\;g\circ f\;$ is $1-1$ doesn't necessarily mean there exist $\;a,a'\in A\;$ ...etc. , as you wrote.
Perhaps a little more accurate could be:
$$\text{Suppose}\;\;\color{red}{g(b)=g(b')}.\;\exists\,a,a'\in A\,\,s.t.\,\,f(a)=b,\,f(a')=b'\;\text{(because $f$ is onto)}\implies$$
$$g(f(a))=\color{red}{g(b)=g(b')}=g(f(a'))\implies g\circ f(a)=g\circ f(a')\implies a=a'\implies b=b'$$
and we're done