1.4Monomorphisms, Epimorphisms

Updated 11 May 2026Chapter (PDF)

[1.4.1]Definition(Monomorphism)#

A function of sets $f: A \to B$ is a monomorphism if, for every set $Z$ and functions $\alpha', \alpha'': Z \to A$ such that $f \circ \alpha' = f \circ \alpha''$ , it follows that $\alpha' = \alpha''$ .

[1.4.2]Theorem(Monomorphisms On Set-Functions Are Equivalent to Injections)#

Suppose $f: A \to B$ is a function of sets. Then $f$ is injective if and only if it is a monomorphism.

Proof.

$(\Longrightarrow):$ Suppose $f$ is injective. Let $Z$ be an arbitrary set and let $\alpha', \alpha'': Z \to A$ be functions such that

f \circ \alpha' = f \circ \alpha''.

Note that $f$ has a left-inverse $g$ , so we may apply it both sides and make use of associativity:

(g \circ f) \circ \alpha' = (g \circ f) \circ a'' \implies \id_A \circ \alpha' = \id_A \circ \alpha'' \implies \alpha' = \alpha''.

Thus $f$ is monomorphic.

$(\Longleftarrow):$ Suppose $f$ is monomorphic. Let $Z$ be an arbitrary set and let $\alpha', \alpha'': Z \to A$ be functions such that

f \circ \alpha' = f \circ \alpha''.

Then it follows that $\alpha = \alpha'$ . For any $z \in Z$ , we have that $\alpha'(z) = a_1 \in A$ and $\alpha''(z) = a_2 \in A$ . Then we have, by equality of the two compositions, that

f(a_1) = f(a_2) \implies a_1 = a_2

for all arbitrary pairs of elements in $A$ . Thus $f$ is injective.

[1.4.3]Definition(Epimorphism)#

A function of sets $f: A \to B$ is a epimorphism if, for every set $Z$ and functions $\beta', \beta'': B \to Z$ such that $\beta' \circ f = \beta'' \circ f$ , it follows that $\beta' = \beta''$ .

[1.4.4]Theorem(Epimorphisms On Set-Functions Are Equivalent to Surjections)#

Suppose $f: A \to B$ is a function of sets. Then $f$ is surjective if and only if it is an epimorphism.

Proof.

$(\Longrightarrow):$ Suppose $f$ is surjective. Let $Z$ be an arbitrary set and let $\beta', \beta'': B \to Z$ be functions such that

\beta' \circ f = \beta'' \circ f.

Note that $f$ has a right-inverse $g$ , so we may apply it both sides and make use of associativity:

\beta' \circ (f \circ g) = \beta'' \circ (f \circ g) \implies \beta' \circ \id_B = \beta'' \circ \id_B \implies \beta' = \beta''.

Thus $f$ is epimorphic.

$(\Longleftarrow):$ Suppose $f$ is not surjective. Let $Z$ be an arbitrary set and let $z_0 \in Z \notin \im(f)$ . Define $\beta', \beta'': B \to Z$ by the following rules

\beta'(\im(f)) = \beta''(\im(f)) \qquad \beta'(z_0) \ne \beta''(z_0).

Then $\beta' \circ f = \beta'' \circ f$ since they agree on $f$ 's image values, but $\beta' \ne \beta''$ . Thus $f$ is not epimorphic, completing a contrapositive proof.

A monomorphism (resp. epimorphism) is denoted by $\hookrightarrow$ (resp. $\twoheadrightarrow$ ).

[1.4.5]Remark(Product Decomposition)#

Observe the decomposition of the Cartesian product of sets $A \times B$ in terms of inclusions (monomorphisms) and projections (epimorphisms) as follows.

The inclusion maps $i_b$ (resp. $i_a$ ) map a set to the product by $i_b(a) = (a, b)$ (resp. $i_a(b) = (a, b)$ )—they are obviously monomorphic. Moreover, the projection maps $\pi_A$ (resp. $\pi_B$ ) project a product onto a set by $\pi_A((a, b)) = a$ (resp. $\pi_B((a, b)) = b$ )—they are obviously epimorphic.

We now reach the canonical decomposition, an extremely important decomposition that yields a natural isomorphism of sets (and perhaps other structures).

[1.4.6]Remark(Canonical Decomposition)#

Let $f: A \to B$ be a map of sets from $A$ to $B$ . Define $\sim$ be an equivalence relation on $S$ such that

a \sim b \iff f(a) = f(b).

Note that the map sending an element $a \in A$ to its equivalence class $[a]_\sim$ is epimorphic—we denote it $\pi: A \twoheadrightarrow A/{\sim}$ . Note $f$ decomposes as follows.

In words, the map $a \mapsto f(a)$ can be decomposed canonically as follows:

Take each $a \in A$ to its equivalence class $[a]_\sim \in A/{\sim}$ . Note the canonical projection $\pi: A \twoheadrightarrow A/{\sim}$ is an epimorphism.
Take each equivalence class $[a]_\sim \in A/{\sim}$ to the image of any element of the equivalence class $f(a) \in \im(f)$ . This is justified by all elements of $[a]_\sim$ mapping to the same element in $\im(f)$ . This is an isomorphism of sets $\stackrel{\sim}{f}: A/{\sim} \stackrel{\sim}{\to} \im(f)$ .
Finally, embed the elements of $\im(f)$ into $B$ by the inclusion map $i: \im(f) \hookrightarrow B$ .

The lengthy description of this process effectively serves as proof.