Existence of a Set Function Axiom of Choice

95 Views Asked by At

I have the following problem. Let $A$ be a set and $B\neq\emptyset$ be a proper subset. Prove the existence of a function $f:A\to A$ such that $f\circ f=f$ and $\text{im}~f=B$.

In the case where $A$ is finite, it suffices to pick a point of $B$, say $x$, and define $f(A)=\{x\}$ while $f_{|B}=\text{id}_{B\to B}$. If $A$ is infinite, I do not know if this argument carries over, as I am very unfamiliar with all possible uses of Axiom of Choice.

Thanks for any advice!

1

There are 1 best solutions below

1
On BEST ANSWER

The axiom of choice is absolutely not needed here.

We only need to choose one element from $b$, since it's not empty we can do it. And the axiom of choice is not needed. Then simply map everything not in $B$ to $b$, while keeping the rest in place.

This is similar to the problem of inversing functions. The axiom of choice is needed in order to construct an injective inverse to a surjective function, but it is not needed in order to construct a surjective inverse to an injective function. (Well, between non-empty sets anyway.)

And this problem is a particular use for the above construction. Here the injection function is $f(b)=b$ from $B$ into $A$.