In Set, you form the pullback of f: A → C and g: B → C. Which set is the pullback?
AA × B — the Cartesian product of A and B
BA ∩ B — the intersection of A and B as subsets of C
C{(a, b) ∈ A × B | f(a) = g(b)} — pairs that agree over C
DA ∪ B — all elements of A and B combined
The pullback is the FIBER PRODUCT: it consists of all pairs (a, b) from A × B for which f(a) = g(b) — elements that 'match' over C. This is smaller than A × B (which ignores f and g entirely). Option B (A ∩ B) only gives the fiber product in the special case where A and B are both subsets of C with f and g being inclusion maps. The fiber product naturally models database joins: pairs of records related through a common foreign key.
Question 2 Multiple Choice
A commutative square P → A, P → B, A → C, B → C satisfies f∘p₁ = g∘p₂. Is P necessarily the pullback of f and g?
AYes — any commutative square over f and g is a pullback by definition
BNo — P is the pullback only if it also has the universal property: any other commutative cone Q factors uniquely through P
CYes — in Set, every commutative square is automatically a pullback
DNo — P is the pullback only if the square is also a pushout
Commutativity is necessary but NOT sufficient for a pullback. The pullback requires the UNIVERSAL PROPERTY: P must be the 'most efficient' solution — every other commutative cone over the cospan must factor uniquely through P. For example, any proper subset of the actual fiber product with restricted maps forms a commutative square but fails the universal property because some Q cannot factor through it. This is the most common misconception about pullbacks.
Question 3 True / False
The pushout in topology allows you to build new spaces by gluing two spaces along a common subspace, and the resulting space carries the quotient topology.
TTrue
FFalse
Answer: True
The pushout in Top of the cospan B ←^f C →^g A is exactly the gluing construction: take the disjoint union B ⊔ A and quotient by identifying f(c) ~ g(c) for each c ∈ C. The quotient topology on this set is precisely the topology that makes it the pushout in the category of topological spaces — continuous maps out of the pushout correspond bijectively to pairs of continuous maps out of A and B that agree on C. Attaching a disk to a circle along its boundary is the canonical example.
Question 4 True / False
Pullbacks and pushouts are categorically dual, so reversing most arrows in any specific pullback square in a category typically yields a valid pushout square in that same category.
TTrue
FFalse
Answer: False
Pullbacks and pushouts are categorically dual — the definition of one is obtained by reversing all arrows in the other. But duality is a statement about the RELATIONSHIP BETWEEN DEFINITIONS across all categories, not a guarantee that reversing a specific diagram in a specific category produces the dual construction in that same category. In Set, the pullback and pushout of the same data are generally different sets. A diagram that is a pullback square will not in general be a pushout square in the same category.
Question 5 Short Answer
Explain the universal property of a pullback and why a commutative square alone is not sufficient to define one.
Think about your answer, then reveal below.
Model answer: A pullback of f: A → C and g: B → C is an object P with morphisms p₁: P → A and p₂: P → B satisfying f∘p₁ = g∘p₂ (commutativity), PLUS the universal property: for any other object Q with morphisms q₁: Q → A and q₂: Q → B satisfying f∘q₁ = g∘q₂, there exists a UNIQUE morphism h: Q → P such that p₁∘h = q₁ and p₂∘h = q₂. Commutativity alone is insufficient because many objects can complete a commutative square — the pullback is the unique one through which all others factor uniquely. It is the 'tightest' completion of the cospan.
The universal property encodes the pullback as the categorical intersection — it captures exactly the information common to A and B over C, no more and no less. Any 'larger' commutative completion (like all of A × B) fails the universal property because the factoring map is not unique. Any 'smaller' subobject fails because some Q cannot factor through it.