Homework 1 soln - Solution PDF

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Problem Set #1: Selected Solutions M367K: Topology I These “selected solutions” are mostly a few hints and outlines of arguments. You will want to imitate the mathematical writing in the book more than the writing here, which often is a hint rather than a solution. Problems in Munkres Section 2 1. (a) For a subset B0 ⊂ B we have   f −1 (B0 ) = a ∈ A : f (a) ∈ B0 .   Hence if a0 ∈ A0 , then since f (a0 ) ∈ f (A0 ) it follows that a0 ∈ f −1 f (A0 ) . Therefore   A0 ⊂ f −1 f (A0 ) . If f is injective, and if a ∈ A satisfies f (a) ∈ f (A0 ) it follows that there exists a0 ∈ A0 such that f (a0 ) = f (a). But then by injectivity a = a0 and so in particular a ∈ A0 . Therefore, f −1 (f (A0 )) ⊂ A0 . Thus, under the hypothesis that f is injective, we have   A0 = f −1 f (A0 ) .   (b) If a ∈ A satisfies a ∈ f −1 (B0 ), then f (a) ∈ B0 . Thus f f −1 (B0 ) ⊂ B0 . If f is surjective and   b ∈ B0 , choose a0 ∈ A such that f (a0 ) = b0 . Then a0 ∈ f −1 (B0 ) from which b0 ∈ f f −1 (B0 ) .   Thus, under the hypothesis that f is surjective, we have f f −1 (B0 ) = B0 . 5. (a) If g : B → A is a left inverse to f , and a, a′ ∈ A satisfy f (a) = f (a′ ), then a = g(f (a)) = g(f (a′ )) = a′ . (b) f : {1} → {a, b} defined by f (1) = a. (c) f : {a, b} → {1} defined by f (a) = f (b) = 1. (d) Yes. The function f : {1, 2} → {a, b, c} which maps 1 7→ a, 2 7→ b has two left inverses which differ only in the image of c. Similarly, the function f : {a, b, c} → {1, 2} which maps a 7→ 1, b 7→ 2, c 7→ 2 has two left inverses which differ only in the image of 2. (e) We must show that for a given b ∈ B there is a unique a ∈ A such that f (a) = b. The right inverse shows existence: f (h(b)) = b. The left inverse shows uniqueness: if a, a′ ∈ A satisfy f (a) = f (a′ ) = b, then a = g(f (a)) = g(f (a′ )) = a′ .

Section 3 3. We must be able to prove aCa for every a ∈ A, where A is the set on which C is a relation. But we do not know every a ∈ A is related to an element of A. 4. (a) Reflexive: f (a) = f (a). Symmetric: f (a0 ) = f (a1 ) iff f (a1 ) = f (a0 ). Transitive: f (a0 ) = f (a1 ) and f (a1 ) = f (a2 ) implies f (a0 ) = f (a2 ). (b) Define f¯: A∗ → B which maps the equivalence class containing a ∈ A to f (a). You must check that f¯ is well-defined, injective, and surjective. I leave it to you—see me if there is any question. Other Problems 1. Argue by contradiction. Write a putative rational number x satisfying x2 = 3 as x = p/q where p, q are integers with no common factors. Then 3q 2 = p2 . This shows that p2 is divisible by 3, and so it follows that p is divisible by 3. (I am relying on very elementary facts about integers here.) So write p = 3p′ for some integer p′ . Then we find q 2 = 3(p′ )2 , and now the same argument shows that q is divisible by 3. But this contradicts the assumption that p and q do not share a common factor. 2. A hint: Take a square and an inscribed circle and map that disk to the interior of the square. Do so by projecting from the center, mapping each line segment emanating from the center by a homothety (stretch). 3. Apply Problem 4 in Section 3: Map the nonzero vectors in R3 to the unit vectors by multiplying by a positive constant....