Questions: Isomorphisms in Categories

Question 1

Let f: X → Y be a continuous bijection between topological spaces. Is f necessarily an isomorphism in the category Top?

• A. Yes — a bijective morphism is always an isomorphism in any category
• B. Yes — continuity plus bijectivity is sufficient for a homeomorphism
• C. No — f is an isomorphism only if its inverse f⁻¹ is also continuous
• D. No — only surjective maps are isomorphisms in Top

Answer: (C)

This is the canonical example showing that 'bijective morphism' ≠ 'isomorphism' outside Set. A homeomorphism requires f to be a continuous bijection *whose inverse is also continuous*. A standard counterexample: let f: [0,1) → S¹ be the map f(t) = e^{2πit}. This is a continuous bijection, but f⁻¹ is not continuous (the preimage of an open arc near f(0) is not open in [0,1)). So f is bijective and continuous but not a homeomorphism — not an isomorphism in Top.

Question 2

In a poset category where objects are elements of a partially ordered set and there is a unique morphism a → b whenever a ≤ b, which morphisms are isomorphisms?

• A. All morphisms — since there is at most one morphism between any two objects, each morphism is trivially invertible
• B. Only identity morphisms — because if a ≤ b and b ≤ a then a = b, so both morphisms exist only when they are the same object
• C. No morphisms — posets have no invertible structure
• D. Any morphism a → b where a and b are comparable elements

Answer: (B)

For f: a → b to be an isomorphism, we need g: b → a such that g∘f = idₐ and f∘g = id_b. In the poset category, a morphism b → a exists only if b ≤ a. Combined with a ≤ b, antisymmetry gives a = b. So the only isomorphisms are identity morphisms (a = b), where the morphism is the identity. This shows that isomorphisms capture the categorical notion of 'the same,' and in a poset, two distinct elements are never 'the same' even if they are comparable.

Question 3

The inverse of an isomorphism f: A → B, when it exists, is unique.

• T. True
• F. False

Answer: True

Suppose g and h are both inverses of f, meaning g∘f = idₐ, f∘g = id_B, h∘f = idₐ, f∘h = id_B. Then: h = h∘id_B = h∘(f∘g) = (h∘f)∘g = idₐ∘g = g. The proof uses only associativity of composition and the identity laws — the axioms of any category. This uniqueness is what makes the notation f⁻¹ well-defined.

Question 4

In any category, a morphism that is bijective on the underlying sets of objects is an isomorphism.

• T. True
• F. False

Answer: False

This is only true in Set. In other categories, morphisms carry additional structure, and the inverse must preserve that structure. In Top, a bijective continuous map need not have a continuous inverse (see homeomorphism counterexample). In a poset category, bijectivity between underlying sets is not even the relevant concept — morphisms represent order relations, not set-maps. The categorical definition of isomorphism — existence of a two-sided inverse in the category — is the correct standard, and it requires the inverse to be a morphism in the same category.

Question 5

Why isn't bijectivity of the underlying set-function sufficient to guarantee an isomorphism in all categories? Give an example where they come apart.

Model answer: Morphisms in a category can carry structure beyond being set-maps. An isomorphism requires a two-sided inverse *that is itself a valid morphism in the category* — it must preserve whatever structure the morphisms represent. In Top, morphisms are continuous maps, so the inverse must also be continuous. A continuous bijection whose inverse is discontinuous is not an isomorphism even though the underlying set-map has an inverse. Example: f: [0,1) → S¹, f(t) = e^{2πit} is a continuous bijection but not a homeomorphism.

The key insight is that 'isomorphism' is a *categorical* notion, defined by composition equations g∘f = id and f∘g = id in the given category. The inverse g must itself be a morphism — it must satisfy the category's requirements on morphisms. Bijectivity only guarantees a set-theoretic inverse exists; it says nothing about whether that inverse is a valid morphism. This is why the categorical definition, which only mentions composition and identities, is more fundamental than any element-based notion.