MHB Show Im(gf)=Im(g) When f is Onto (Wondering)

[mathmari]

Hey!

Let $f:B\rightarrow C$ and $g:C\rightarrow D$.
I want to show that Im(gf)=Im(g), when f is onto. I have done the following:

Let $x\in \text{Im}g$.
Then $\exists y\in C$ such that $g(y)=x$.
Since $f$ is onto, we have that $\exists b\in B$ such that $f(b)=y$.
Then $g(f(b))=x\Rightarrow x=(gf)(b)$.
So, $x\in \text{Im}(gf)$.

Let $x\in \text{Im}(gf)$ then $\exists z\in B$ such that $x=(gf)(z)=g(f(z))$.
Since $f$ is onto, we have that $\forall c\in C \ \ \exists b\in B$ such that $f(b)=c$.
We choose $c=f(z)$ and then we have that $x=g(c)\Rightarrow x\in \text{Im}(g)$.

Is my last step correct? I am not really sure.. (Wondering)

 

[Fallen Angel]

Hi mathmari,

There is no need of $f$ to be onto to prove $Im(gf)\subseteq Im(g)$ since, by definition, $f(z)\in C$

 

[mathmari]

There is no need of $f$ to be onto to prove $Im(gf)\subseteq Im(g)$ since, by definition, $f(z)\in C$

Let $x\in \text{Im}(gf)$ then $\exists z\in B$ such that $x=(gf)(z)=g(f(z))$.
We have that $f(z)\in C$, say $c:=f(z)$, then we have that $x=g(c)\Rightarrow x\in \text{Im}(g)$.

Is this correct? Or didn't you mean it so? (Wondering)

 

[Fallen Angel]

Yes, that's correct.

 

[mathmari]

Thank you very much! (Yes)