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For a surjective function $f:B\to A$ and a function $g:B\to C$, there is a unique function $h:A\to C$ with $g=h\circ f$ if and only if $g$ is constant on each fiber of a function of $f$. Define $h(a)$ to be the common value on $f^{-1}(a)$; it is well-defined and unique because that fiber is nonempty. This uniquely specified value requires no simultaneous arbitrary choices.

For a concrete example, restrict the codomain of the finite function above to its image of a function $I=\{0,1,2\}$, making $f:A\to I$ surjective. Define $g(a)=g(b)=5$, $g(c)=7$ and $g(d)=9$. Then $h(y)=2y+5$ on $I$ gives $g=h\circ f$: knowing the output of $f$ is enough to compute $g$. If instead $g(a)\ne g(b)$, no such $h$ can exist, because $h(f(a))=h(f(b))$ must be the same value.

The surjectivity assumption supplies uniqueness on the full target. If the original unused target value three were retained, the factorization would impose no condition on $h(3)$, and different extensions could agree on every actual input. Thus factorization through a surjection formalizes when a compressed description retains all information needed for a later calculation.

// Target: combinatorics.bigb

For a function $f:X\to Y$ and $y\in Y$, its fiber of a function over $y$ is the set of all inputs producing that one output:Here $f^{-1}(y)$ is shorthand for a preimage of a singleton set, rather than a claim that an inverse function exists. A fiber can contain several inputs or be empty. It describes exactly which distinctions the function forgets.

For the finite function under image and preimage of a function, the fibers over $0,1,2,3$ are $\{a,b\}$, $\{c\}$, $\{d\}$ and the empty set, respectively. The nonempty fibers form a partition of a set: they are disjoint sets whose union is the domain of a function. Every input belongs to the fiber over its unique output, and two fibers with distinct output labels cannot share an input. A function is injective exactly when every fiber has at most one member; it is surjective exactly when every member of its specified codomain has a nonempty fiber.

More generally, $f^{-1}(U)$ is the union of the fibers over all $y\in U$. The relation $x\sim x^{\prime }$ defined by $f(x)=f(x^{\prime })$ is an equivalence relation, and its equivalence classes are precisely the nonempty fibers. The resulting quotient set is naturally in bijection with the image of a function. This links an input partition to the output values actually reached, without choosing one representative from each class.

Fibers are useful when an output is known but the input is not uniquely determined. A solution set $g^{-1}(\{0\})$ is one fiber, and factorization through a surjection explains when another function depends only on the retained output. Quantum hidden subgroup problem algorithms use functions whose fibers are cosets, giving the same idea a structured form in group theory.

Choose a right inverse $s:A\to B$ of the surjection $f$, so $f(s(a))=a$, and defineFor each $a$, the element $b=s(a)$ then witnesses both $f(b)=a$ and $g(b)=h(a)$. As in part (a), arbitrary simultaneous choices use the axiom of choice. If $A$ is empty, surjectivity and the existence of $f:B\to A$ force $B$ to be empty, and the unique empty function $h:A\to C$ already meets the requirement.

Suppose $g$ is constant on each fiber of a function of $f$. Since every fiber is nonempty, $h(a)$ must be its common $g$-value: any witness $b$ in that fiber gives that value. This defines exactly one $h$ and yields $g=h\circ f$, an instance of factorization through a surjection.

Conversely, suppose a fiber contains $b,b^{\prime }$ with $f(b)=f(b^{\prime })=a_0$ but $g(b)\ne g(b^{\prime })$. Starting with any admissible $h$, define $h_1,h_2$ to agree with it away from $a_0$, but put $h_1(a_0)=g(b)$ and $h_2(a_0)=g(b^{\prime })$. Both remain admissible, witnessed at $a_0$ by $b$ and $b^{\prime }$ respectively, yet they are distinct. HenceThe distinction is important: the original existence property asks only for one witness in each fiber. Only when $g$ is fiberwise constant does that property force $g=h\circ f$ for every $b$.