Products by maps: a first universal property

इस पर आधारित

8 अनिवार्य पूर्वापेक्षाएँ (बिल्कुल नीचे तक)

आप सीखेंगे

• Given f : X → A and g : X → B, construct the unique h : X → A × B with π_A ∘ h = f and π_B ∘ h = g.
• Draw the commuting diagram for this property and explain what 'exactly one' means.
• Show that any set P with maps to A and B having the same property is in bijection with A × B by a unique bijection compatible with the maps.

परिभाषा

• The projections π_A : A × B → A and π_B : A × B → B send (a, b) to a and to b respectively.
• The pairing ⟨f, g⟩ : X → A × B of f : X → A and g : X → B sends x to (f(x), g(x)).
• A universal property describes an object by how it relates, through maps, to all other objects of the same kind.

प्रमेय

• For every set X, functions X → A × B correspond one-to-one with pairs of functions X → A and X → B.
• If a set P with maps p : P → A and q : P → B has the same property, there is exactly one bijection P → A × B compatible with the maps.
• For finite sets this matches counting: |A × B|^|X| = |A|^|X| · |B|^|X|.

आम भ्रांतियाँ

• Thinking the universal property defines a different object from A × B; it describes A × B by its behaviour, up to a unique compatible bijection.
• Forgetting the uniqueness part of the property.
• Believing the set of ordered pairs is the only possible product; any set with the property serves equally well.

आगे ले जाता है

वैकल्पिक खोजें

• Products, coproducts, initial and terminal objects In any category, a product of A and B is defined by exactly this property; for groups it is the direct product, for topological spaces the product topology.विस्मय पथ →
• The Yoneda lemma An object is determined up to isomorphism by the maps into it from all other objects; the Yoneda lemma makes this precise.
• ← Sets
• ← Functions

अधिगम पथ

• Accelerated secondary progression

आगे पढ़ने के लिए

• F. William Lawvere and Stephen H. Schanuel, Conceptual Mathematics: A First Introduction to Categories (1997)