Composite of Epimorphisms is Epimorphism

Theorem

Let:

$\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$

$\struct {S_2, *_1, *_2, \ldots, *_n}$

$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

be algebraic structures.

Let:

$\phi: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$

$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

be epimorphisms.

Then the composite of $\phi$ and $\psi$ is also an epimorphism.

Proof

From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.

From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surjection.

An epimorphism is a surjective homomorphism.

Hence $\psi \circ \phi$ is an epimorphism.

$\blacksquare$

Sources

• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$