PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For $f : A \to B$ and $g : B \to C$ : if both are injective so is $g \circ f$ ; if both are surjective so is $g \circ f$ ; if $g \circ f$ is injective so is $f$ ; and if $g \circ f$ is surjective so is $g$

Statement

Let $f : A \to B$ and $g : B \to C$ . Then $g \circ f : A \to C$ , and

(i) if $f$ and $g$ are injective then so is $g \circ f$ ;
(ii) if $f$ and $g$ are surjective then so is $g \circ f$ ;
(iii) if $g \circ f$ is injective then so is $f$ ;
(iv) if $g \circ f$ is surjective then so is $g$ .

In particular a composite of bijections is a bijection.

Facts & Assumptions

Given: functions $f : A \to B$ and $g : B \to C$ .

[L1]

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

[L2]

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

[L3]

$f$ is bijective if it is both injective and surjective (Injection, surjection, bijection).

[L4]

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

[L5]

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).

[L6]

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

[L7]

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

Proof

technique · direct

1.1

$g \circ f : A \to C$ : its domain is $f^{-1}[\operatorname{dom} g] = f^{-1}[B]$ , which is $A$ because $\operatorname{dom} f = A$ and $\operatorname{ran} f \subseteq B$ ; and its values $g(f(a))$ lie in $\operatorname{ran} g \subseteq C$ .

L4L5L6L7

2.1

Claim (i): assume $f$ and $g$ injective and let $a, a' \in A$ with $g(f(a)) = g(f(a'))$ . Injectivity of $g$ gives $f(a) = f(a')$ , and injectivity of $f$ gives $a = a'$ .

L1L4step 1.1

2.2

Claim (ii): assume $f$ and $g$ surjective and let $c \in C$ . Surjectivity of $g$ gives $b \in B$ with $g(b) = c$ , and surjectivity of $f$ gives $a \in A$ with $f(a) = b$ ; then $g(f(a)) = c$ .

L2L4step 1.1

2.3

Claim (iii): assume $g \circ f$ injective and let $a, a' \in A$ with $f(a) = f(a')$ . Then $g(f(a)) = g(f(a'))$ , so $a = a'$ .

L1L4step 1.1

2.4

Claim (iv): assume $g \circ f$ surjective and let $c \in C$ . There is $a \in A$ with $g(f(a)) = c$ , and $f(a)$ is an element of $B$ , so $c$ is a value of $g$ .

L2L4L5step 1.1

3.1

Claims (i) to (iv) are established; combining (i) and (ii) gives that a composite of bijections is injective and surjective, hence a bijection.

L3step 2.1step 2.2step 2.3step 2.4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Bijection, injection and surjection (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)
Function composition (Wikipedia) (standard reference, not scraped)

For f:A→Bf : A \to Bf:A→B and g:B→Cg : B \to Cg:B→C: if both are injective so is g∘fg \circ fg∘f; if both are surjective so is g∘fg \circ fg∘f; if g∘fg \circ fg∘f is injective so is fff; and if g∘fg \circ fg∘f is surjective so is ggg