Problem Sheet 2 Solutions
... Suppose X is Hausdorff and let x, y ∈ X with x 6= y. Then there exist open sets U , V with x ∈ U , y ∈ V and U ∩ V = ∅. Since U and V are open and non-empty, we see that X \ U and X \ V are both finite and hence the union (X \ U ) ∪ (X \ V ) is finite. But (X \ U ) ∪ (X \ V ) = X \ (U ∩ V ) = X \ ∅ ...
... Suppose X is Hausdorff and let x, y ∈ X with x 6= y. Then there exist open sets U , V with x ∈ U , y ∈ V and U ∩ V = ∅. Since U and V are open and non-empty, we see that X \ U and X \ V are both finite and hence the union (X \ U ) ∪ (X \ V ) is finite. But (X \ U ) ∪ (X \ V ) = X \ (U ∩ V ) = X \ ∅ ...
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... Here, the notation indicates that we’re talking about ∂n (it’s the only one of these maps that I could apply to such an object, so leaving the n out is unambiguous), and that (P0 , . . . , P̌i , . . . , Pn ) = (P0 , . . . , Pi−1 , Pi+1 , . . . , Pn ). It’s σ with the ith entry removed. So we have a ...
... Here, the notation indicates that we’re talking about ∂n (it’s the only one of these maps that I could apply to such an object, so leaving the n out is unambiguous), and that (P0 , . . . , P̌i , . . . , Pn ) = (P0 , . . . , Pi−1 , Pi+1 , . . . , Pn ). It’s σ with the ith entry removed. So we have a ...
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... Given a topological space X, the diagonal embedding, or diagonal map of X into X × X (with the product topology) is the map ...
... Given a topological space X, the diagonal embedding, or diagonal map of X into X × X (with the product topology) is the map ...
USC3002 Picturing the World Through Mathematics
... {B } . Lindelöff’s theorem 6.13 implies X is Lindelöf hence Theorem 8.9 implies X is normal. Therefore for each pair (i, j ) N N with Bi B j there exists f : X [0,1] such that f ( Bi ) 0 and f ( X \ B j ) 1. - why? These functions can be ...
... {B } . Lindelöff’s theorem 6.13 implies X is Lindelöf hence Theorem 8.9 implies X is normal. Therefore for each pair (i, j ) N N with Bi B j there exists f : X [0,1] such that f ( Bi ) 0 and f ( X \ B j ) 1. - why? These functions can be ...
Garrett 02-15-2012 1 Harmonic analysis, on R, R/Z, Q , A, and A
... S 1 so that E contains no non-trivial subgroups of G. Using the b be the open compactness of G itself, let U ⊂ G b : f (G) ⊂ E} U = {f ∈ G Since E is small, f (G) = {1}. That is, f is the trivial b for compact G. homomorphism. This proves discreteness of G ...
... S 1 so that E contains no non-trivial subgroups of G. Using the b be the open compactness of G itself, let U ⊂ G b : f (G) ⊂ E} U = {f ∈ G Since E is small, f (G) = {1}. That is, f is the trivial b for compact G. homomorphism. This proves discreteness of G ...
Cell-Like Maps (Lecture 5)
... Proposition 2. Let f : X → Y be a homeomorphism of finite CW complexes with the property that for every cell e ⊆ Y , the inverse image f −1 e is a union of cells. Then f is a simple homotopy equivalence. We postpone the proof; we will discuss a stronger result at the end of this lecture. Let us inst ...
... Proposition 2. Let f : X → Y be a homeomorphism of finite CW complexes with the property that for every cell e ⊆ Y , the inverse image f −1 e is a union of cells. Then f is a simple homotopy equivalence. We postpone the proof; we will discuss a stronger result at the end of this lecture. Let us inst ...
1. (a) Let X be a topological space and γ 0,γ1 : [0
... (b) What does it mean to say that two topological spaces are homotopy equivalent? ...
... (b) What does it mean to say that two topological spaces are homotopy equivalent? ...
Quiz 1 solutions
... A = B, then A = B. The statement is false. Let X = R with the usual topology. Also, let A = [0, 1] and B = (0, 1). Then A = B = [0, 1] although A 6= B. PROBLEM 5 (10 points): Let X and Y be topological spaces and let B be a basis for Y . Show that a map f : X → Y is continuous if and only if for eve ...
... A = B, then A = B. The statement is false. Let X = R with the usual topology. Also, let A = [0, 1] and B = (0, 1). Then A = B = [0, 1] although A 6= B. PROBLEM 5 (10 points): Let X and Y be topological spaces and let B be a basis for Y . Show that a map f : X → Y is continuous if and only if for eve ...
1.6 Smooth functions and partitions of unity
... which is given by (p, u) 7→ ((ψ ◦ ϕ−1 )(p), D(ψ ◦ ϕ−1 )p u), which is smooth. Therefore we obtain a topology and smooth structure on all of T M (by defining W to be open when W ∩ π −1 (U) is open for every U in an atlas for M; all that remains is to verify the Hausdorff property, which holds since p ...
... which is given by (p, u) 7→ ((ψ ◦ ϕ−1 )(p), D(ψ ◦ ϕ−1 )p u), which is smooth. Therefore we obtain a topology and smooth structure on all of T M (by defining W to be open when W ∩ π −1 (U) is open for every U in an atlas for M; all that remains is to verify the Hausdorff property, which holds since p ...
