Every function $f : A \to B$ factors as $f = \iota \circ \bar f \circ \pi$ with $\pi : A \to A/{\ker f}$ the surjective quotient map, $\bar f : A/{\ker f} \to f[A]$ the bijection $[a] \mapsto f(a)$, and $\iota : f[A] \to B$ the inclusion injection; $\bar f$ is the only function making the factorisation hold

Given: a function $f : A \to B$, with $\sim$ denoting $\ker f$.

$\ker f$ is a set, is a relation on $A$, and is an equivalence relation on $A$ (For a function $f : A \to B$ the relation $\ker f := \{\,(a,a') \in A \times A : f(a) = f(a')\,\}$ is an equivalence relation on $A$).

$\pi$ is a surjective function $A \to A/{\sim}$ with $\pi(a) = [a]$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $\pi : A \to A/{\sim}$, and let $f : A \to B$. There is a function $g : A/{\sim} \to B$ with $g \circ \pi = f$ if and only if $a \sim a'$ implies $f(a) = f(a')$; and such a $g$ is then unique).

there is a function $g : A/{\sim} \to B$ with $g \circ \pi = f$ if and only if $f(a) = f(a')$ whenever $a \sim a'$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $\pi : A \to A/{\sim}$, and let $f : A \to B$. There is a function $g : A/{\sim} \to B$ with $g \circ \pi = f$ if and only if $a \sim a'$ implies $f(a) = f(a')$; and such a $g$ is then unique).

when such a $g$ exists it is unique, and it satisfies $g([a]) = f(a)$ for every $a \in A$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $\pi : A \to A/{\sim}$, and let $f : A \to B$. There is a function $g : A/{\sim} \to B$ with $g \circ \pi = f$ if and only if $a \sim a'$ implies $f(a) = f(a')$; and such a $g$ is then unique).

$A/{\sim} \;:=\; \{\, [a] \;:\; a \in A \,\}$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$).

$f$ is injective (one-to-one) if $f(x) = f(y)$ implies $x = y$, for all $x, y \in A$ (Injection, surjection, bijection).

$f$ 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).

$b \in R[A]$ holds if and only if $(a,b) \in R$ for some $a \in A$ (The image $R[A]$ and the preimage $R^{-1}[B]$ of a set under a relation).

$\Delta_A$ 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 $f : A \to B$).

$g \circ f$ 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 $f : A \to B$).

$f = g$ if and only if $\operatorname{dom} f = \operatorname{dom} g$ and $f(x) = g(x)$ for every $x \in \operatorname{dom} f$ (Functions $f$ and $g$ are equal if and only if $\operatorname{dom} f = \operatorname{dom} g$ and $f(x) = g(x)$ for every $x$ in that common domain).

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

$(a,b) \in \Delta_A$ 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 $\in_A\, = \{\,(a,b) \in A \times A : a \in b\,\}$).

$T \circ (S \circ R) = (T \circ S) \circ R$ ($T \circ (S \circ R) = (T \circ S) \circ R$, $(S \circ R)^{-1} = R^{-1} \circ S^{-1}$, $(R^{-1})^{-1} = R$, $\operatorname{dom}(R^{-1}) = \operatorname{ran} R$, and $\Delta_B \circ R = R = R \circ \Delta_A$ for a relation $R$ from $A$ to $B$).

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$, proper subset $x \subsetneq y$, and the separation notation $\{\, z \in x : \varphi(z) \,\}$).

$a \sim b$ if and only if $[a] = [b]$ (The equivalence classes of an equivalence relation are nonempty, cover $A$, and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).