Suppose that $\;f_1,\;f_2:\;A\to B$ such that $f_1$ is injective and $f_2$ is surjective.
I was trying to find out whether there exists $f_3:\;A\to B$ such that $f_3$ is bijective. Is it possible to construct it from $f_1$ and $f_2$?
I tried proving the statement non-constructively via a cardinality argument, but ended up hand-waving a lot (i.e., since the cardinality is the same, there exists a bijection, a bit of circular reasoning). But, I'd be more interested in a construction based argument, as I can't think of one.
Best Answer
Assuming choice, yes.
The surjection can be reversed, so you have two injections and the Cantor–Bernstein finishes the job.
Note that if $B$ is empty then $A$ is empty, since there is an injection from $A$ to $B$.
Without choice, however, there is an injection from $\Bbb R$ into $[\Bbb R]^\omega$, the set of countably infinite subsets of $\Bbb R$, and there is a surjection as well (fix a bijection between $\Bbb R$ and $\Bbb{R^N}$, and then map every sequence to its range, or the natural numbers if the range is finite).
But a theorem of Sierpinski show that if there is a bijection between $\Bbb R$ and $[\Bbb R]^\omega$ then there are sets without the property of Baire and non-measurable sets. Since it is consistent that all sets have the property of Baire, or all sets are Lebesgue measurable, in such models there is no bijection between the two sets.
(There are other examples of this sort, e.g. $\Bbb R$ and $\Bbb{R/Q}$.)