# Is the inverse of a one-to-one, total function itself one-to-one and total?

The problem, from Boolos and Jeffrey's Computability and Logic, states the following definitions:

1. "If a function $f(a)$ (from $A$ to $B$) is defined for every element $a$ of $A$, then it is called total"

2. "A correspondence between sets $A$ and $B$ is a one-to-one, total function from $A$ onto $B$."

3. "Two sets are said to be equinumerous if and only if there is a correspondence between $A$ and $B$."

Then, it asks the reader to show that if $A$ is equinumerous with $B$, then $B$ is equinumerous with $A$. So far, the work I've done is to show that, for a function $f : A \rightarrow B$, there exists a function $f^{-1} : B \rightarrow A$ that is onto (which follows from $f$ being total), and one-to-one (which follows from $f$ existing). However, it remains to be shown that $f^{-1}$ is total. This seems to require that $f$ is onto, which doesn't seem to be stated by the problem. I would appreciate any hints on how to show that $f^{-1}$ must be total, given the constraints.

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Did you miss the "from $A$ onto $B$"? – Robert Israel Jul 1 '12 at 21:55
@Arturo: I suppose that total means that the domain is the entire $A$ since computability often deals with partial functions, it makes sense to distinguish total functions. – Asaf Karagila Jul 1 '12 at 21:56
@RobertIsrael, does that imply that $f$ is onto? I'm not familiar with the terminology or convention. – arbn Jul 1 '12 at 21:57
@arbin: Look carefully at the penultimate word in condition 2. You notice that it is onto and not merely "to"? That's what tells you that $f$ is onto. – Arturo Magidin Jul 1 '12 at 21:57

It is common to write that the function is "onto $B$" to mean that it is surjective (onto). Otherwise, we would say that a function $f$ is a [total] function "from $A$ to $B$".
So when it says that it is a total one-to-one, function from $A$ onto $B$ it is telling you (i) the domain is all of $A$; (ii) the function is one-to-one; and (iii) the image is all of $B$.