Helly I have the following problem.
We have $f:T\rightarrow S$ a map an $\mathfrak{R}$ a relation on $S$. We define a relation $\mathfrak{P}$ with $$x~\mathfrak{P}~y \Leftrightarrow f(x)~\mathfrak{R}~f(y)$$Now we assume that $\mathfrak{R}$ is an equivalence relation and we define $$g:T/\mathfrak{P}\rightarrow S/\mathfrak{R};~~~~~T\supset B\mapsto A\subset S~~~s.t.~~~f(B)\subset A$$I just know that $g$ is always injective and I know want to show that if $f$ is surjective then $g$ is bijective.
So my Idea was the following. I remarked that it is enough to show that if $f$ is surjective then $g$ is surjective. I wanted to assume that $g$ is not surjective then $$\exists A\in S/\mathfrak{R}~~~s.t.~~~\forall B\in T/\mathfrak{P}~~~g(B)\neq A \Rightarrow f(B)\not\subset A$$ In my opinion this shows that $f$ is not surjective and thus the claim follows by contraposition, but I'm not really sure if this is correct so?
Could someone take a look and help me?
Thank you a lot.
This is not enough; how do you know there is actually some element $s\in A$ such that $f(t)\neq s$ for all $t\in T$?
To say something here you really need to get your hands on what the elements of $S/\mathfrak R$ look like: an element here has the form $A=[s]$ for $s\in S$, the equivalence class.
How do you proceed for a proof? Suppose $f$ is surjective. Now if $A\in S/\mathfrak R$ then $A=[s]$ for some $s\in S$. But because $f$ is surjective take some $t\in T$ such that $f(t)=s$, and from here show that you can take $B=[t]\in T/\mathfrak P$ and get $g(B)=A$.