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.

Question

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

Answer 1

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

Answer 2

0) Want to show : For $b \not =b'$ : $g(b)\not =g(b')$, i.e $g $ injective.

1) Let $b \not = b'$;

2) Since $f$ is surjective there are $a,a'$, $a \not = a'$, s.t.

$f(a)=b$, and $f(a')=b'$.

3) $g(f(a))=g(b)$, and $g(f(a'))=g(b')$.

4)Since $g \circ f$ is injective , and $a \not =a'$

$g(b)=g(f(a)) \not = g(f(a')) = g(b')$ , and we are done.