# kt3k/math-exercises-notes

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# 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.