Composite of Surjections is Surjection
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Theorem
A composite of surjections is a surjection.
That is:
- If $g$ and $f$ are surjections, then so is $g \circ f$.
Proof
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be surjections.
Then:
\(\ds \forall z \in S_3: \exists y \in S_2: \, \) | \(\ds \map g y\) | \(=\) | \(\ds z\) | Definition of Surjection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in S_1: \, \) | \(\ds \map f x\) | \(=\) | \(\ds y\) | Definition of Surjection |
By definition of a composite mapping:
- $\map {g \circ f} x = \map g {\map f x} = \map g y = z$
Hence $g \circ f$ is surjective.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations: $\text{(i)}$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text I$: Sets and Functions: Factoring Functions
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (2)$
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.4$: Functions: Theorem $\text{A}.4$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $2$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $2$: Maps and relations on sets: Exercise $2 \ \text{(b)}$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(d)}$
- 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text A$: Set Theory: Functions