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!
Best Answer
$f^{-1}[C]$ is also countable as a $2$-$1$ preimage of a countable set.
So $A$ is the union of a countable set $f^{-1}[C]$ and a set in $A \setminus f^{-1}[C] \simeq B\setminus C$ (by $f$ restricted). As any two countably infinite sets are equinumerous, $B \simeq A$ is then obvious.