real analysis - Finding one-to-one correspondence between two infinite sets

Suppose $f:A\rightarrow B$ is a surjective map of infinite sets. Let $C$ be a countable subset of $B$. Suppose that $f^{-1}(y)$ has two elements if $y\in C$, and one element if $y\in B-C$. Show that there is a one-to-one correspondence between $A$ and $B$.

So I thought to define a function $\phi:B\rightarrow A$. It's given that for $y\in B-C$, $f^{-1}(y)$ is one-to-one so we know for those elements we are fine. The only thing to do is find a correspondence for the elements in $C$. But since we know that there are two elements in $A$ corresponding to each of these elements in $C$. How can this possibly be one-to-one?

Thanks for any help offered!

Answer

The trick is to exploit the countability of C. So there is an element 1 of C, element 2 of C, etc. Rewrite f as follows: Leave f on B - C alone. For each pair of elements of A each mapped to the same element of C, arbitrarily designate one of them as 'odd' and the other as 'even'. Then remap them by parity - ie, change the mapping from the i'th pair in A to the i'th element of C to map the 'odd' element to the (2*i-1)'th element of C and the 'even' element to the (2*i)'th element of C.