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We have the sets $X,Y$ and $f:X\to Y, \ g: Y\to X $

Prove if $f$ has a right inverse function: $f\circ g=id_Y$ $\iff$ $f$ is onto $Y$ (surjective).

$\Leftarrow \forall y\in Y : \exists x\in X$ then $f(y)=x$ I need to show that $f\circ g(x)=x$ but I don't know how.

I have no clue on to start in the other direction.

BTW, I found below similar question but they're either not quite the same as this one or I didn't understand the answers:

Left inverse iff injective; right inverse iff surjective

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    $\begingroup$ For "$\Leftarrow$" you need to choose a preimage for every $y\in Y$, to do this you absolutely need the axiom of choice! $\endgroup$ – Christoph May 11 '14 at 13:34
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  • For $\boxed{\Leftarrow}$: assume $f$ is surjective. Then, for all $y\in Y$ there exists $x_y\in X$ such that $f(x_y)=y$. Define $g\colon Y\to X$ to be the function which maps each $y\in Y$ to such $x\in X$ (if there is more than one $x$, then the function $g$ maps $y$ to one of them chosen in an arbitrary way. This excludes the possibility that $g$ map $y$ to two distinct values, in which case it wouldn't be a function). It follows that $$ \forall y\in Y,\quad f\circ g(y) = f( g(y) ) = f(y_x) = y $$ and $f\circ g=id_Y$.

  • For $\boxed{\Rightarrow}$: assume $f\colon X\to Y,g\colon Y\to X$ are such that $f\circ g=id_Y$. Then for each $y\in Y$, $x_y\stackrel{\rm def}{=} g(y)\in X$ is a preimage of $y$ by $f$, as $f(x_y)=f\circ g(y) = id_Y(y)=y$. Hence $f$ is surjective.

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  • $\begingroup$ Why did you subtract the $idy(y)$ ? $\endgroup$ – shinzou May 11 '14 at 13:36
  • $\begingroup$ Oh, it should be an $=$. $\endgroup$ – Clement C. May 11 '14 at 13:37
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$\Leftarrow)$ For all $y\in Y$ there's $x\in X$ such that $f(x)=y$ so define $g$ by $$g\colon Y\rightarrow X,\quad y\mapsto x$$ and we have $$f\circ g=id_Y$$

$\Rightarrow)$ Let $y\in Y$ so $f\circ g(y)=id_Y(y)=y$ so let $x=g(y)\in X$ hence we have $$\exists x=g(y)\in X,\quad f(x)=y$$ so $f$ is onto.

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  • $\begingroup$ It's all by definitions. Thanks. $\endgroup$ – shinzou May 11 '14 at 13:49
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 May 11 '14 at 13:50
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    $\begingroup$ I think your proof should emphasize the fact that if $g(y)$ has the possibility of being mapped to more than one $x \in X$, then one $x$ is chosen arbitrarily to guarantee that $g$ is well-defined (unambiguous). $\endgroup$ – Jeremiah Dunivin Dec 5 '15 at 22:50

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