If there is a mapping of $B$ onto $A$, then $2^{|A|} \leq 2^{|B|}$

Question

If there is a mapping of $B$ onto $A$, then $2^{|A|} \leq 2^{|B|}$. [Hint: Given $g$ mapping $B$ onto $A$, let $f(X)=g^{-1}(X)$ for all $X \subseteq A$]

I follow the hint and obtain the function $f$. If $f$ is injective, then the statement is proven.

Question: Why does $g^{-1}$ exist in the first place? How do we know $g$ is injective? The hint given seems a bit weird.

Can anyone explain to me?

Answer 1

Question: Why does $g^{−1}$ exist in the first place?

It exists for all maps. Here, it does not denote the inverse, but the pre-image map,

$$g^{-1}(X) = \{b\in B : g(b)\in X\}.$$

Now use the surjectivity of $g$ to deduce the injectivity of $g^{-1}$.

Answer 2

If there exists a function $f:B \longrightarrow A$ such that $f$ is onto then $\lvert B \rvert \geq \lvert A \rvert$. And this means that there exists an injective function $g: A \longrightarrow B$. Now as we want to see that $2^{\lvert A \rvert} \leq 2^{\lvert B \rvert}$ it's enough to define an injective function $h: \mathcal P \left( {A} \right) \longrightarrow \mathcal P \left( {B} \right)$. Such that for $C \subseteq A$, $ h(C) :=${$ g(x) \lvert x \in C$}. Now it's not to hard to prove that this function is well defined and injective.