elementary set theory - Bijective function

Recently I was wondering if we need the Axiom of Choice in order to find an inverse function given an bijective funcion:

If $f:A \rightarrow B$ is bijective we mean that $f$ is injective and surjective. Assume that $f:A \rightarrow B$ is bijective. I want to define $f^{-1}: B \rightarrow A$ s.t. $f \circ f^{-1} = 1_B$ and $f^{-1} \circ f = 1_A$. For each $b \in B$ let $X_b := \{x \in X: f(x) = b \}$. By surjectivity each $X_b$ is non-empty. But not knowing wether $A$ or $B$ are finite I need AC in order to select from each $X_b$ an (in fact unique) element $x_b$ and defining $f^{-1}(b) := x_b$ for each $b \in B$.

Is AC necassary ?