FALSE: assuming ZF is consistent, ZF proves that every surjection $f : A \to B$ has a right inverse $g : B \to A$ with $f \circ g = \Delta_B$

Given: ZF is consistent, and the claim above.

If ZF is consistent, then ZF does not prove the Axiom of Choice (Cohen 1963, Cohen 1963: ZF does not prove the Axiom of Choice ‡). This is an external result, established by forcing and quoted rather than proved here.

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f(x) = b$ (Injection, surjection, bijection).

We write $f : A \to B$, and say $f$ is a function from $A$ to $B$, when $f$ is a function with $\operatorname{dom} f = A$ and $\operatorname{ran} f \subseteq B$ (A function is a relation $f$ with $(a,b) \in f$ and $(a,c) \in f$ implying $b = c$; $f : A \to B$, the value $f(a)$, domain and codomain).

an element of $\prod_{i \in I} A_i$ is a function with domain $I$ that takes its value at each index inside the member carried by that index (The product $\prod_{i \in I} A_i := \{\, f : I \to \bigcup_{i \in I} A_i \ \mid\ f(i) \in A_i \text{ for every } i \in I \,\}$).

if $A_j = \varnothing$ for some $j \in I$ then $\prod_{i \in I} A_i = \varnothing$ ($\prod_{i \in \varnothing} A_i = \{\varnothing\}$; if $A_j = \varnothing$ for some $j \in I$ then $\prod_{i \in I} A_i = \varnothing$; and for $I = \{j\}$ the evaluation $f \mapsto f(j)$ is a bijection $\prod_{i \in I} A_i \to A_j$).

$g \circ f$ is a function, $\operatorname{dom}(g \circ f) = f^{-1}[\operatorname{dom} g]$, and $(g \circ f)(x) = g(f(x))$ for every $x$ in that domain (If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{-1}[\operatorname{dom} g]$ and $(g \circ f)(x) = g(f(x))$ there; $\Delta_A$ is a function with $\Delta_A(a) = a$; and $f \circ \Delta_A = f = \Delta_B \circ f$ for $f : A \to B$).

$\Delta_A$ is a function with $\operatorname{dom} \Delta_A = A$ and $\Delta_A(a) = a$ for every $a \in A$ (If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{-1}[\operatorname{dom} g]$ and $(g \circ f)(x) = g(f(x))$ there; $\Delta_A$ is a function with $\Delta_A(a) = a$; and $f \circ \Delta_A = f = \Delta_B \circ f$ for $f : A \to B$).

$(a,b) \in \Delta_A$ holds if and only if $a = b$ and $a \in A$ (The identity relation $\Delta_A = \{\,(a,b) \in A \times A : a = b\,\}$ and the membership relation $\in_A\, = \{\,(a,b) \in A \times A : a \in b\,\}$).

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$).

An indexed family with index set $I$ is a function $A$ with $\operatorname{dom} A = I$ (An indexed family $(A_i)_{i \in I}$ is a function with domain $I$; $\{A_i : i \in I\}$ is its range).

For any parameters $\bar p$ and any set $x$, there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $\varphi(z,\bar p)$ holds (The Axiom Schema of Separation: for each formula $\varphi$, $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$).

$\operatorname{dom} R := \{\, a : \exists b\ (a,b) \in R \,\}, \qquad \operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\}$ (Relation, $\operatorname{dom} R$, $\operatorname{ran} R$, $\operatorname{fld} R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$").

$(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ($(a,b) = (c,d)$ if and only if $a = c$ and $b = d$).

$z \in \bigcup_{i \in I} A_i$ holds if and only if $z \in A_i$ for some $i \in I$ ($\bigcup_{i \in I} A_i := \bigcup \{A_i : i \in I\}$, and $\bigcap_{i \in I} A_i := \bigcap \{A_i : i \in I\}$ for $I \neq \varnothing$).

An equivalent formulation of the Axiom of Choice is that a product of nonempty sets is nonempty: if $X_i \ne \varnothing$ for every $i \in I$, then $\prod_{i \in I} X_i \ne \varnothing$ (The Axiom of Choice).