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README.md

README.md

categories

Every non-obvious statement in each of the examples appearing below should be treated as an exercise

Nice way to present exercises 👍

Exercises

1.3

  • Let 𝒞 be an category and X an object in 𝒞.
  • Let f and g be identities of X.
  • Because f is an identify, f ⚬ g = g. --- (1)
  • Because = is symmetric and (1), g = f ⚬ g. --- (2)
  • Because g is an identify, f ⚬ g = f. --- (3)
  • Because (2), (3), and = is transitive, g = f. QED

1.4

(1) Sets is a category

  • Let X, Y, Z, W be objects in Sets.
  • Let f: X -> Y, g: Y -> Z, h: Z -> W be morphisms.
  • For any x in X, ((h ⚬ g) ⚬ f)(x) = (h ⚬ g)(f(x)) = h(g(f(x))) = h((g ⚬ f)(x)) = (h ⚬ (g ⚬ f))(x), which means (h ⚬ g) ⚬ f = h ⚬ (g ⚬ f).
  • Therefore the composition satisfies the axiom of Categories, and therefore Sets is a category.

(2) sets is a category

  • The same argument holds.

(3) Groups is a category

  • Let X, Y, Z be objects in Groups.
  • Let f: X -> Y, g: Y -> Z be morphisms.
  • Let x, y be elements of X.
  • (g ⚬ f)(x * y) = g(f(x * y)) - (1) because of the composition definition.
  • g(f(x * y)) = g(f(x) * f(y)) - (2) because f is a homomorphism.
  • g(f(x) * f(y)) = g(f(x)) * g(f(y)) - (3) because g is a homomorphism.
  • g(f(x)) * g(f(y)) = (g ⚬ f)(x) * (g ⚬ f)(y) - (4) because of the composition definition.
  • Because of (1), (2), (3), and (4), (g ⚬ f)(x * y) = (g ⚬ f)(x) * (g ⚬ f)(y).
  • Therefore g ⚬ f is a homomorphism, which means the composition definition is closed among homomophisms, and therefore Groups is a category.
  • (For the composition axiom, the same arguments as Sets holds.)

(4) groups is a category

  • The same arguments as (3) hold.

(5) Ab is a category

  • The same arguments as (3) hold.

(6) ab is a category

  • The same as (3).

(7) Vect_k is a category

  • Let X, Y, Z be vector spaces over k.
  • Let f: X -> Y, g: Y -> Z be linear maps.
  • Let x ∈ X.
  • Let a ∈ k.
  • (g ⚬ f)(ax) = g(f(ax)) --- by definition of g ⚬ f
  • = g(af(x)) --- because f is a linear map.
  • = ag(f(x)) --- because g is a linear map.
  • = a(g ⚬ f)(x) --- by definition of g ⚬ f
  • g ⚬ f is a linear map from X to Z.
  • The composition is closed among linear maps, which means Vect_k forms a category. QED.

(8) vect_k is a category

  • The same as (7).

(9) Top is a category

  • Let X, Y, Z be topological spaces.
  • Let f: X -> Y, g: Y -> Z be continuous maps.
  • Let U ⊂ Z be an open set.
  • (g ⚬ f)^-1(U) = f^-1(g^-1(U)) --- by definition of the inverse map.
  • g^-1(U) is an open set in Y because g is a continuous map.
  • Therefore f^-1(g^-1(U)) is an open set in X because f in a continuous map.
  • This means g ⚬ f is a continuous map and the composition is closed among continous maps.
  • This measn Top forms a category. QED.

1.5

1st question

  • Let X be a topological space.
  • Let U, V, W in Op(X) such that U ⊆ V and V ⊆ W.
  • Let a: U -> V, b: V -> W, c: U -> W.
  • b ⚬ a equals inevitably c because there is no other choice.

2nd question

  • Let U, V, W, X in Op(X) such that U ⊆ V ⊆ W ⊆ X.
  • Let f: U -> V, g: V -> W, h: W -> X, k: U -> X.
  • (h ⚬ g) ⚬ f = k = h ⚬ (g ⚬ f) because there is no other choice.

1.6

The inverse is unique

  • Let 𝒞 be a category, f: X -> Y be a morphism, g, h is inverse of f.
  • g = g ⚬ id_Y = g ⚬ (f ⚬ h) = (g ⚬ f) ⚬ h = id_X ⚬ h = h. QED.

The converse

  • Let X = { 0 }, Y = { 0, 1 } in Sets.
  • Let f: X -> Y, f(0) = 0 and g: Y -> X, g(0) = g(1) = 0.
  • Then g ⚬ f = id_X and this is invertible by itself.
  • However f isn't invertible and g either isn't invertible.
  • Therefore the converse is not true.