Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Using Schroeder-Bernstein Theorem. Assume there exists a 1-1 function $f:X→Y$ and another 1-1 function $g:Y→X$. If we define $f^{−1}(y)=x$, then $f^{−1}$ is a 1-1 function from $f(X)$ onto $X$. Also, we can define the 1-1 function $g^{-1}: g(X)→Y$. Follow the steps to show that there exists a 1-1, onto function $h:X→Y$.

I'm not sure if I am over thinking this or missing the obvious, but if $f$ maps $X$ to $Y$, will $g$ map $f(X)$ to $X$ or all of $Y$ to $X$? And is this why $g^{-1}$ maps $g(X)$ to $Y$?

share|improve this question
The arrows indicate mappings into the codomain, not onto it. So $g$ maps both $f(X)$ and $Y$ into $X$, but not necessarily onto $X$. (If it did, there would be nothing left to do, you would just choose $h=g^{-1}$.) However, $g^{-1}$ as an inverse of a function defined on $Y$ really maps $g(X)$ onto $Y$. –  Lukas Geyer Oct 6 '12 at 21:49
That clarifies things, but can you define g(X) please? Because it confused me, I thought it would be g(Y). –  Alti Oct 6 '12 at 22:15
My mistake (or a mistake in the statement of the problem that I did not catch.) You are correct, it should always be $g(Y)$, since $g(X)$ is not defined (unless $X \subseteq Y$.) –  Lukas Geyer Oct 6 '12 at 22:43
Thank you, I thought that too but doubted myself. –  Alti Oct 6 '12 at 22:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.