2
$\begingroup$

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.

  • 0
    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".2012-11-29

1 Answers 1

6

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.