As the title says, I would like to prove that $f\colon X\to Y$ bijective $\Longleftrightarrow$ $f$ has an inverse function.



Let $f$ be bijective. That means $\forall y\in Y\exists! x\in X: f(x)=y$. Define $f^{-1}\colon Y\to X, y\mapsto x$, then $f^{-1}(y)=f^{-1}(f(x))=x$.


Let $f^{-1}\colon Y\to X$ be the inverse function of $f\colon X\to Y$. Surjectivity of $f$: Let $y\in Y$. Then $f^{-1}(y)\in X$ and $y=f(f^{-1}(y))$, so there exists a $x\in X: f(x)=y$. So $f$ is surjective.

Injectivity of $f$: Consider $f(x)=f(y)$. Then $x=f^{-1}(f(x))=f^{-1}(f(y))=y$. So $f$ is injective.

So $f$ is injective and surjective so it is bijective.

That's my proof. Is it okay?

  • $\begingroup$ your proof seems good! $\endgroup$ – the_candyman May 4 '14 at 18:10
  • $\begingroup$ @mathfemi you just add in the first implication that $f\circ f^{-1}(y)=y$ to conclude that $f^{-1}$ you defined is the inverse of $f$. $\endgroup$ – Walid Taamallah May 4 '14 at 18:20

The $\implies$-part is incomplete/informal to me. But if you say this is your sketch of the proof then I will say it's correct. For incomplete/informal, I meant:

$$\textrm{To prove}\ g:Y\to X\ \textrm{is an inverse of}\ f:X\to Y\ \textrm{, you have to show, by definition, that:}\ \\ f\circ g=I_Y\ \textrm{and}\ g\circ f=I_X.$$

But you didn't show the former one. A proof is short doesn't mean it's better.

About the $\Longleftarrow$-part, it can be made more clear:

$$\textrm{Let}\ f^{-1}(y)=x.\\ f\ (\ f^{-1}(y))=f(x).\ \textrm{But}\ f^{-1}\ \textrm{is inverse means the LHS is}\ y,\ \textrm{so}\ f\ \textrm{is surjective.}$$

And your remaining part is ok to me.

I won't promise that if you properly apply a definition then your proof must become shorter. But I can promise you it must be more clear.

The reference of my words: http://math.colorado.edu/~kstange/has-inverse-is-bijective.pdf.


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