# Is there a way to see if a function is invertible without proving it is onto and one-to-one?

For any function?

Right now, I try to find the values of x and y for the function to see if it is one-to-one, but it doesn't work for some of the more complex and unusual functions.

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This is question is too general. In any rate, you can find and inverse. The theorem says that if it is onto and one to one, then an inverse exists, so it gives you a tool to prove existence of an inverse without actually finding it. – Lior B-S Nov 29 '12 at 7:33
Find an inverse. In the parts of mathematics that I enjoy this is sometimes, perhaps often, easier than proving that a function is onto and one-to-one. – Qiaochu Yuan Nov 29 '12 at 7:37
isn't finding an inverse harder sometimes though? – mathnoob Nov 29 '12 at 7:43
there is finding the x and y values and finding a counter examples i guess... – mathnoob Nov 29 '12 at 7:44
By proving it is invertible, you have proven that a function is injective (one-to-one) and surjective (onto), so the answer to your question is "no". – robjohn Nov 29 '12 at 15:43

Let $f:A\to B$ be a function.

If both $A$ and $B$ are finite and $f$ is injective then it has an inverse.

If both $A$ and $B$ are finite and $f$ is surjective then it has an inverse.

If both $A$ and $B$ are vector spaces, both hava the same finite dimension, $f$ is a linear transformation and $f$ is injective then it has an inverse.

If both $A$ and $B$ are vector spaces, both hava the same finite dimension, $f$ is a linear transformation and $f$ is surjective then it has an inverse.

If both $A$ and $B$ are vector spaces and $f$ is a linear transformation that maps a basis to a basis then it has an inverse.

If $A$ and $B$ are posets and $f$ is strictly monotonic and surjective then it has an inverse.

If $f$ is both left and right cancelable (in the category of sets) then it has an inverse.

If $A,B\subseteq \mathbb R^n$ are nice enough domains, $f$ is differentiable on $A$, the Jacobian of $f$ at some $a\in A$ is an invertible matrix, $A$ is small enough, and $f$ is onto then it has an inverse.

There are more such criteria of course from various areas of mathematics.

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