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If $f\colon X \to Y$ is bijective then there exists $g \colon Y \to X$ such that (i) $g\circ f = \operatorname{Id}_X$ and (ii) $f\circ g = \operatorname{Id}_Y$.

Conversly show that if (i) is true then $f$ is one to one and if (ii) is true then $f$ is onto.

I'm struggling to go about solving this. I understand that the identity function is one where when you compose it with anything it doesn't change it. I've been working along the lines of $g(f(x))=f^{-1}(f(x))$ and therefore $g(x)=f^{-1}(x)$ and so $g\colon Y\to X$? Any help at all would be appreciated.

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    $\begingroup$ This involves the use of axiom of choice. $\endgroup$ – Troy Woo Oct 5 '14 at 19:20
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Proposition 1: Let $f : X \to Y$ be a function. Then $f$ is injective iff $f$ has a left inverse, that is, there exists a function $g : Y \to X$ such that $g \circ f = Id_{X}$.

You want a proof of the $[\Longleftarrow]$ part.

Proof of the $[\Longleftarrow]$ part: Let $x_1, x_2 \in X$ s.t. $f(x_1) = f(x_2)$. Hence $(g \circ f)(x_1) = g(f(x_1)) = g(f(x_2)) = (g \circ f)(x_2)$. But $g \circ f = Id_X$ therefore $x_1 = x_2$, which shows $f$ is injective. $\blacksquare$

Proposition 2: Let $f:X\to Y$ be a function. Then $f$ is surjective iff $f$ has a right inverse, that is, there exists a function $g:Y \to X$ such that $f \circ g = Id_Y$.

You want a proof of the $[\Longleftarrow]$ part.

Proof of the $[\Longleftarrow]$ part: Let $y \in Y$. Then $(f \circ g)(y) = Id_Y(y) = y$. But $(f \circ g)(y) = f(g(y))$ also, therefore $x:=g(y) \in X$ is an element s.t. $f(x) = y$, which shows $f$ is surjective. $\blacksquare$

Note that these results are very general. Therefore there is not much choice on how to proceed to prove them...

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If there exists $g:Y\to X$ such that $g\circ f=id_X$ then $f$ is injective because:

$$f(x_1)=f(x_2)\implies x_1=id_X(x_1)=(g\circ f)(x_1)=(g\circ f)(x_2)=id_X(x_2)=x_2.$$

If there exists $g:Y\to X$ such that $f\circ g=id_Y$ then $f$ is surjective because for any $y\in Y$ there exists $x\in X$ such that $f(x)=y.$ Indeed, $x=g(y):$

$$y=id_Y(y)=(f\circ g)(y)=f(g(y)).$$

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