2.2 Properties and types of morphisms

Morphisms are the arrows that connect objects in categories. They come in different flavors: monomorphisms, epimorphisms, and isomorphisms. Each type has unique properties that define how objects relate to one another.

Understanding these morphism types is crucial for grasping category theory. They help us see how mathematical structures connect and transform, giving us a powerful lens to view relationships across various mathematical domains.

Properties of Morphisms

Types of morphisms

• Monomorphism (monic morphism or left-cancellative morphism)
  • A morphism $f: A \rightarrow B$ is a monomorphism if it is left-cancellative
  • For any object $C$ and morphisms $g_1, g_2: C \rightarrow A$, if $f \circ g_1 = f \circ g_2$, then $g_1 = g_2$
  • Injective functions are monomorphisms in the category of sets
  • The inclusion map from a subgroup to its parent group is a monomorphism in the category of groups
  • The inclusion map from a subspace to its parent vector space is a monomorphism in the category of vector spaces
• Epimorphism (epic morphism or right-cancellative morphism)
  • A morphism $f: A \rightarrow B$ is an epimorphism if it is right-cancellative
  • For any object $C$ and morphisms $g_1, g_2: B \rightarrow C$, if $g_1 \circ f = g_2 \circ f$, then $g_1 = g_2$
  • Surjective functions are epimorphisms in the category of sets
  • The quotient map from a group to its quotient group is an epimorphism in the category of groups
  • The projection map from a vector space to its quotient space is an epimorphism in the category of vector spaces
• Isomorphism
  • A morphism $f: A \rightarrow B$ is an isomorphism if there exists an inverse morphism $g: B \rightarrow A$
  • $g \circ f = id_A$ and $f \circ g = id_B$, where $id_A$ and $id_B$ are the identity morphisms of objects $A$ and $B$, respectively
  • Bijective functions are isomorphisms in the category of sets
  • The isomorphism between a group and its isomorphic copy is an isomorphism in the category of groups
  • The isomorphism between a vector space and its isomorphic copy is an isomorphism in the category of vector spaces

Left vs right invertibility

• Left invertibility
  • A morphism $f: A \rightarrow B$ is left invertible if there exists a morphism $g: B \rightarrow A$ such that $g \circ f = id_A$
  • If $f$ is left invertible, it is a monomorphism because it is left-cancellative
  • The left inverse $g$ is not necessarily unique
• Right invertibility
  • A morphism $f: A \rightarrow B$ is right invertible if there exists a morphism $h: B \rightarrow A$ such that $f \circ h = id_B$
  • If $f$ is right invertible, it is an epimorphism because it is right-cancellative
  • The right inverse $h$ is not necessarily unique

Relationships among morphism types

• Isomorphisms are both monomorphisms and epimorphisms
  • If $f: A \rightarrow B$ is an isomorphism, then it is both left and right invertible
  • Being left invertible implies $f$ is a monomorphism, and being right invertible implies $f$ is an epimorphism
• In some categories, monomorphisms and epimorphisms are not necessarily isomorphisms
  • The inclusion map from the integers to the rational numbers is both a monomorphism and an epimorphism in the category of rings
  • However, it is not an isomorphism because there is no inverse map from the rational numbers to the integers

Isomorphisms as mono-epimorphisms

• To prove an isomorphism is a monomorphism:
  1. Let $f: A \rightarrow B$ be an isomorphism with inverse $g: B \rightarrow A$
  2. Suppose $h_1, h_2: C \rightarrow A$ are morphisms such that $f \circ h_1 = f \circ h_2$
  3. Compose both sides with $g$ to get $g \circ (f \circ h_1) = g \circ (f \circ h_2)$
  4. By associativity, $(g \circ f) \circ h_1 = (g \circ f) \circ h_2$
  5. Since $g \circ f = id_A$, we have $h_1 = h_2$, proving $f$ is a monomorphism
• To prove an isomorphism is an epimorphism:
  1. Let $f: A \rightarrow B$ be an isomorphism with inverse $g: B \rightarrow A$
  2. Suppose $k_1, k_2: B \rightarrow C$ are morphisms such that $k_1 \circ f = k_2 \circ f$
  3. Compose both sides with $g$ to get $(k_1 \circ f) \circ g = (k_2 \circ f) \circ g$
  4. By associativity, $k_1 \circ (f \circ g) = k_2 \circ (f \circ g)$
  5. Since $f \circ g = id_B$, we have $k_1 = k_2$, proving $f$ is an epimorphism