Suppose there is a function that is bijective $f: X \to Y$, and it has an inverse $f^{-1}:Y\to X$. Verify that $f^{-1}(f(x))=x, \forall x\in > X$.
Here is my attempt:
If $f^{-1}(f(x))=x, \forall x\in X$, then by applying $f$ to both sides: $f(f^{-1}(f(x))=f(x)$
By surjectivity: for each $f(x)\in y, \exists f^{-1}(f(x))\in X$ Such that the equation is true.
For an arbitrary $y=f(x)$, there exists an arbitrary $x=f^{-1}(f(x))$, then if the proposition is false, this equation should hold:
$f(y)\neq f(x)$
Which is true iff $y\neq x$, but that contradicts the subjectivity statement.
To conclude: My proof is a mess and I believe that I have used circular logic, more so where the bolded text is.
How would I go about proving this?
Let $y=f^{-1}(f(x))$. Then $f(y)=f(f^{-1}(f(x))=f(x)$ by definition of $f^{-1}$. [$f(f^{-1}u)=u$ for all $u$ in the range of $f$. I am taking $u$ to be $f(x)$]. Since $f$ is one-to-one this implies $y=x$ as required, .