# Proof by Contradiction to show that if $f^{-1}$ exists, $f$ must be onto

Use proof by contradiction to prove that if $f^{-1}$ exists, then $f$ must be onto where $f:A→B$.

Proof:

I think the contradiction of the theorem would be: if $f$ exists then $f^{-1}$ must be onto where $f:A→B$.

To me, this contradiction statement seems to be true. Since $f$ maps all values in $A$ to $B$ that means for $f^{-1}$ all values are mapped back to $A$ from $B$. This would be defined as 'onto' since all values in set $A$ are used.

I suppose what is still unknown is whether $f^{-1}$ is defined for all values of $B$. Is this fact what makes the contradiction from being true? I know the contradiction should be false for the original theorem to be true.

And help would be greatly appreciated!

• Sorry I meant contradiction as written in the original problem statement. I'll fix my question. – James Taylor Apr 13 '15 at 4:12

The statement is false, assuming that "$f^{-1}$ exists" means "$f^{-1}$ is a function" and not $f^{-1}:B\to A$; for example the identity function $f:\Bbb N\to\Bbb R$ defined by $f(x)=x$ has an inverse function $f^{-1}:\Bbb N\to\Bbb N$ as $f(x)=x$, but it is not surjective onto $\Bbb R$.
If we assume that $f^{-1}$ is a function with domain $B$, then it is true, because for any $b\in B$ we have $f(f^{-1}(b))=b$, so $f$ is surjective onto $B$.
It is not necessary to use proof by contradiction, but if you wanted to write it that way, it would go something like this: Suppose $f:A\to B$ and $f^{-1}:B\to A$, yet $f$ is not surjective. Then there is some $b\in B$ such that $f(a)\ne b$ for all $a\in A$. Pick $a=f^{-1}(b)$, which is defined since $b$ is in the domain of $f^{-1}$; then $f(f^{-1}(b))=b$, a contradiction.