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I have a problem I need help with: If $X,Y$ are $\neq$ non-empty sets and f: X$\rightarrow$Y a transformation. Prove that the following statements are equivalent: (i) f is surjective (ii) There is a transformation g: Y$\rightarrow$ X so that f $\circ$ g = $I_Y$ ( Identity of Y).

My solution is the following: Let c $\in$ C be arbitrary. So that there exists an a with c = f(g(a)). Let's define b := g(a) so that there is a b $\in$ B with g(b) = f(g(a)) = c. Any hints or solutions guiding to the right direction I much appreciate.

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  • $\begingroup$ I also wanted to ask if I have to proof the statement from the other direction? $\endgroup$ – SAINT May 11 '18 at 19:07
  • $\begingroup$ Who is $C$? Maybe $Y$. $\endgroup$ – paf May 11 '18 at 19:08
  • $\begingroup$ I took a random letter. $\endgroup$ – SAINT May 11 '18 at 19:42
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You have to prove an equivalence, hence you need to prove two assertions: (i)$\Rightarrow$(ii) and (ii)$\Rightarrow$(i).

(i)$\Rightarrow$(ii) Hint: Take any element $y$ of $Y$ and try to construct $g(y)$ such that $f(g(y))=y$.

(ii)$\Rightarrow$(i) Hint: Take any element $y$ of $Y$ and use $g$ to find $x\in X$ such that $f(x) = y$.

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  • $\begingroup$ (i) $\Rightarrow$ (ii) Let x $\in$ X be arbitrary. Then there exists an $y$ with y = f(g(x)). Define y := g(x) with y $\in$ Y with with g(x) = f(g(x)) = y $\endgroup$ – SAINT May 11 '18 at 23:36
  • $\begingroup$ (ii) $\Rightarrow$ (i) Let y $\in$ Y be arbitrary. Then there exists an x with x = f(g(y)). Define x := g(y) with x $\in$ X to get g(y) = f(g(y)) = x. Is that a correct answer? $\endgroup$ – SAINT May 11 '18 at 23:43
  • $\begingroup$ First part: what is $g(x)$? $g$ isn't defined yet (you have to do it) and even if it were defined, it would have no meaning since $g$ can only take elements of $Y$ whereas $x\in X$. Hint: you must use the surjectivity of $f$. 2nd part: be clear on your choice of $x$: is it $f(g(y))$ or $g(y)$? Moreover, $g(y) \ne f(g(y))$ and anyway, you didn't check that $f(x) = y$. $\endgroup$ – paf May 12 '18 at 16:23

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