one-one and onto proofs

f onto implies that there exists an inverse function

No.

Consider $f:\mathbb R \to [0, \infty)$ via $f(x) = x^2$ or $g:\mathbb R \to [-1,1]$ via $f(x) = \sin x$. These functions are onto but they are not one-to-one. For any $f(x) = y$ there maybe two solutions to $x$ (if $x$ is a solution so is $-x$) and $\sqrt{x}$ is not an inverse function. And if $g(x) =y$ there will be infinitely many solutions (if $x$ is a solution, then so is $x + k2\pi$). $\arcsin x$ is not an inverse function.

I realize mathematicians are not consistant with notation and that is their fault, not yours. but $f^{-1}(A)$ does not mean an inverse function but .... if $f:X\to Y$ and $A\subset Y$ then $f^{-1}(A)$ means all the elements of $X$ that get mapped do any element of $A$.

Example if $f(x) = x^2$ then $f^{-1}(\{16\}) = \{4,-4\}$ because $4$ and $-4$ are the elements so that $f(x) = 16$. And $f^{-1}(\{16,25\}) = \{4,-4,5,-5\}$ because those are the elements that get mapped to $16$ or $25$.

And $f^{-1}(\{-2\}) = \emptyset$ as nothing gets mapped to $-2$. And $f^{-1}(\{-2, -25, 3, 49\}) = \{\sqrt 3,-\sqrt{3}, 7, -7\}$ because those are all the elements that get mapped to $-2,-25, 3,$ or $49$.

....

So your question.

$f$ being onto means there is $x\in [-1,1]$ so that $f(x) =0$. There may be many of them. There may be an infinite number of them. But ther is one. So $f^{-1}(\{0\})$ is not the empty set.

Now by definition $f^{-1}(\{0\}) = \{x\in [-1,1]| f(x) = 0\}$.

And be definition $f(A) = \{f(x)|x\in A\}$.

So.......

$f(f^{-1}(\{0\})) = f(\{x\in [-1,1]|f(x)=0\}) =$

$\{f(x)|x \in \{x\in[-1,1]|f(x)=0\}\}$

..... well if $x\in \{x\in[-1,1]|f(x)=0\}$ then that means $f(x) = 0$.

So $\{f(x)|x \in \{x\in[-1,1]|f(x)=0\}\}=$

$\{0\}$.

And that is it.