List 6
... 1. Let (X, TX ) and (Y, TY ) be nonempty topological spaces, and let A ⊆ X and B ⊆ Y be proper subsets. If X and Y are connected, show that the complement of (A × B) is connected. 2. Let (X, T) be a topological space, and let (Y, TY ) be a quotient space of X (that is, Y = X/ ∼ for some equivalence ...
... 1. Let (X, TX ) and (Y, TY ) be nonempty topological spaces, and let A ⊆ X and B ⊆ Y be proper subsets. If X and Y are connected, show that the complement of (A × B) is connected. 2. Let (X, T) be a topological space, and let (Y, TY ) be a quotient space of X (that is, Y = X/ ∼ for some equivalence ...
Algebraic topology exam
... Answer eight questions, four from part I and four from part II. Give as much detail in your answers as you can. Part I 1. Prove the Zig-Zag lemma: let 0 C D E 0 be a short exact sequence of chain complexes with the above maps being f: C D, g : D E. Show that there is a long exact sequenc ...
... Answer eight questions, four from part I and four from part II. Give as much detail in your answers as you can. Part I 1. Prove the Zig-Zag lemma: let 0 C D E 0 be a short exact sequence of chain complexes with the above maps being f: C D, g : D E. Show that there is a long exact sequenc ...
Garrett 02-15-2012 1 Harmonic analysis, on R, R/Z, Q , A, and A
... Given non-trivial ψ ∈ (A/k)b, the k-vectorspace k ·ψ inside (A/k)b b ≈ A. Assuming for a moment injects to a copy of k · ψ inside A that the image in A is essentially the same as the diagonal copy of k, (A/k)b/k injects to A/k. The topology of (A/k)b is discrete, and the quotient (A/k)b/k is still d ...
... Given non-trivial ψ ∈ (A/k)b, the k-vectorspace k ·ψ inside (A/k)b b ≈ A. Assuming for a moment injects to a copy of k · ψ inside A that the image in A is essentially the same as the diagonal copy of k, (A/k)b/k injects to A/k. The topology of (A/k)b is discrete, and the quotient (A/k)b/k is still d ...
4.2 Simplicial Homology Groups
... any two points. Let ϕ be a permutation of points {p0 , p1 , . . . , pk }. Then (pϕ(0) , pϕ(1) , . . . , pϕ(k) ) = (sign ϕ)(p0 , p1 , . . . , pk ), where sign ϕ is the parity of the permutation ϕ. ...
... any two points. Let ϕ be a permutation of points {p0 , p1 , . . . , pk }. Then (pϕ(0) , pϕ(1) , . . . , pϕ(k) ) = (sign ϕ)(p0 , p1 , . . . , pk ), where sign ϕ is the parity of the permutation ϕ. ...
Topological embeddings of graphs in graphs
... NONEMBEDDING EXAMPLE. If E is a graph homeomorphic to a Figure Eight and T is a graph homeomorphic to a Figure Theta, then neither is homeomorphic to a subspace of the other. The proof of this assertion has two parts. One uses the local homology groups at points which are defined in Section VII.1 of ...
... NONEMBEDDING EXAMPLE. If E is a graph homeomorphic to a Figure Eight and T is a graph homeomorphic to a Figure Theta, then neither is homeomorphic to a subspace of the other. The proof of this assertion has two parts. One uses the local homology groups at points which are defined in Section VII.1 of ...
Lecture 5 Notes
... smooth manifold, we’re going to require some compatibility between these coordinate charts for when they overlap. Definition 2.2. Let X be locally Euclidean of dimension n. A smooth atlas A on X is a collection {(U↵ , '↵ )} of coordinate charts satisfying (1) The U↵ ’s cover X, that is ...
... smooth manifold, we’re going to require some compatibility between these coordinate charts for when they overlap. Definition 2.2. Let X be locally Euclidean of dimension n. A smooth atlas A on X is a collection {(U↵ , '↵ )} of coordinate charts satisfying (1) The U↵ ’s cover X, that is ...
(pdf)
... Yet another way of stating this fact is to say that the groups Dn (K) form a subchain complex, with maps d obtained by restriction. ...
... Yet another way of stating this fact is to say that the groups Dn (K) form a subchain complex, with maps d obtained by restriction. ...
k h b c b a q c p e a d r e m d f g n p r l m k g l q h n f
... Solution: Every compact connected surface is homeomorphic to S 2 , a connected sum of copies of P 2 , or a connected sum of copies of T 2 . Moreover, the surfaces in the above list are pairwise non-homeomorphic. The manifold above has 10 triangles, 15 edges, and 6 vertices. Its Euler characteristc i ...
... Solution: Every compact connected surface is homeomorphic to S 2 , a connected sum of copies of P 2 , or a connected sum of copies of T 2 . Moreover, the surfaces in the above list are pairwise non-homeomorphic. The manifold above has 10 triangles, 15 edges, and 6 vertices. Its Euler characteristc i ...
Simplicial Sets - Stanford Computer Graphics
... • If X is a topological space, and A is a contractible subspace of X, then the quotient map X X/A is a homotopy equivalence • Any n-simplex of a simplicial complex is contractible ...
... • If X is a topological space, and A is a contractible subspace of X, then the quotient map X X/A is a homotopy equivalence • Any n-simplex of a simplicial complex is contractible ...
Midterm Exam Solutions
... X/∼, which is connected by hypothesis, giving a contradiction. (4) Consider the set R with the finite-complement topology. (This is the topology in which the nonempty open sets are exactly the sets with finite complement.) Answer each of the following questions about this topological space, and give ...
... X/∼, which is connected by hypothesis, giving a contradiction. (4) Consider the set R with the finite-complement topology. (This is the topology in which the nonempty open sets are exactly the sets with finite complement.) Answer each of the following questions about this topological space, and give ...
Lecture 3 TOPOLOGICAL CONSTRUCTIONS In this lecture, we
... Then we denote the quotient space (X ∪ Y )/∼, where ∼ is the equivalence relation identifying points in each of the sets Fb , b ∈ B, X ∪f Y and say that this space is obtained by attaching (or gluing) Y to X along f . 3.5. Cone, Suspension, and Join (i) Roughly speaking, the cone over a space is obt ...
... Then we denote the quotient space (X ∪ Y )/∼, where ∼ is the equivalence relation identifying points in each of the sets Fb , b ∈ B, X ∪f Y and say that this space is obtained by attaching (or gluing) Y to X along f . 3.5. Cone, Suspension, and Join (i) Roughly speaking, the cone over a space is obt ...
Partitions of Unity
... 15. Definition. A topological space X has covering dimension at most m iff every open cover of X has a refinement with the property that each point of X is contained in at most m + 1 members of the refinement. The covering dimension of X is the least m such that X has covering dimension at most m; i ...
... 15. Definition. A topological space X has covering dimension at most m iff every open cover of X has a refinement with the property that each point of X is contained in at most m + 1 members of the refinement. The covering dimension of X is the least m such that X has covering dimension at most m; i ...
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 ...
Week 5 Term 2
... (draw (3), (4)). Isomorphism of coverings (without fixing basepoints) is just a graph isomorphism preserving labeling and orientation. Note also that characteristic subgroups may be isomorphic without being conjugate. (draw (5),(6)), these are homeomorphic graphs, but not isomorphic as covering spac ...
... (draw (3), (4)). Isomorphism of coverings (without fixing basepoints) is just a graph isomorphism preserving labeling and orientation. Note also that characteristic subgroups may be isomorphic without being conjugate. (draw (5),(6)), these are homeomorphic graphs, but not isomorphic as covering spac ...
MA 331 HW 15: Is the Mayflower Compact? If X is a topological
... (3) (*) Prove that a topological graph G is compact if and only if G has finitely many edges and vertices. (4) (Challenging!) Suppose that X and Y are topological spaces. Let C(X,Y ) be the set of continuous functions X → Y . We give C(X,Y ) a topology T (called the compact-open topology) as follows ...
... (3) (*) Prove that a topological graph G is compact if and only if G has finitely many edges and vertices. (4) (Challenging!) Suppose that X and Y are topological spaces. Let C(X,Y ) be the set of continuous functions X → Y . We give C(X,Y ) a topology T (called the compact-open topology) as follows ...
Prof. Girardi The Circle Group T Definition of Topological Group A
... Let’s look at some nice properties of T. Consider the natural projection π : R T given by π (θ) = [θ]. Then π is continuous since if dR (xn , x) → 0 then dT ([xn ] , [x]) → 0. Following directly from the definition of the quotient topology is that π is an open mapping and that T is Hausdorff. T is ...
... Let’s look at some nice properties of T. Consider the natural projection π : R T given by π (θ) = [θ]. Then π is continuous since if dR (xn , x) → 0 then dT ([xn ] , [x]) → 0. Following directly from the definition of the quotient topology is that π is an open mapping and that T is Hausdorff. T is ...
Topology I – Problem Set Five Fall 2011
... 2. Let Gi be a collection of more than one non-trivial group. Prove that their free product is non-abelian, contains elements of infinite order, and that its center is trivial. 3. Let G , H, G0 , and H 0 be cyclic groups of orders m, n, m0 , and n0 respectively. If G ∗ H is isomorphic to G0 ∗ H 0 th ...
... 2. Let Gi be a collection of more than one non-trivial group. Prove that their free product is non-abelian, contains elements of infinite order, and that its center is trivial. 3. Let G , H, G0 , and H 0 be cyclic groups of orders m, n, m0 , and n0 respectively. If G ∗ H is isomorphic to G0 ∗ H 0 th ...
Notes on wedges and joins
... pointed topological space (Z, z0 ) where Z is the quotient space Z = X t Y /{x0 = y0 }, where t denotes the disjoint union, and z0 is the point obtained by identification of x0 with y0 . The wedge sum is written as (X, x0 ) ∨ (Y, Y0 ), or simply as X ∨ Y . The latter notation is generally misleading ...
... pointed topological space (Z, z0 ) where Z is the quotient space Z = X t Y /{x0 = y0 }, where t denotes the disjoint union, and z0 is the point obtained by identification of x0 with y0 . The wedge sum is written as (X, x0 ) ∨ (Y, Y0 ), or simply as X ∨ Y . The latter notation is generally misleading ...
homework 1
... 1. Prove that a topological manifold M is connected if and only if it is path-connected. Define an equivalence relation on Rn+1 \ {0} by x ∼ y ⇐⇒ y = tx, t ∈ R − {0}. The n-dimensional, real projective space RPn is the quotient of Rn+1 by the above equivalence relation. 2. Prove that RPn is second c ...
... 1. Prove that a topological manifold M is connected if and only if it is path-connected. Define an equivalence relation on Rn+1 \ {0} by x ∼ y ⇐⇒ y = tx, t ∈ R − {0}. The n-dimensional, real projective space RPn is the quotient of Rn+1 by the above equivalence relation. 2. Prove that RPn is second c ...
The No Retraction Theorem and a Generalization
... Proof. We proceed by contradiction. Suppose f : D → S 1 is a retraction from the unit disk to its boundary. Let α, ω be points on the unit circle: α, ω ∈ S 1 . Note that S 1 \ {α, ω} is disconnected, as it is the union of two disjoint relative open sets. Thus, it suffices to show that it is the imag ...
... Proof. We proceed by contradiction. Suppose f : D → S 1 is a retraction from the unit disk to its boundary. Let α, ω be points on the unit circle: α, ω ∈ S 1 . Note that S 1 \ {α, ω} is disconnected, as it is the union of two disjoint relative open sets. Thus, it suffices to show that it is the imag ...
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... by [σ][τ ] = [στ ], where στ means “travel along σ and then τ ”. This gives [(S 1 , 1), (X, x0 )] a group structure and we define the fundamental group of (X, x0 ) to be π1 (X, x0 ) = [(S 1 , 1), (X, x0 )]. In general, the fundamental group of a topological space depends upon the choice of basepoint ...
... by [σ][τ ] = [στ ], where στ means “travel along σ and then τ ”. This gives [(S 1 , 1), (X, x0 )] a group structure and we define the fundamental group of (X, x0 ) to be π1 (X, x0 ) = [(S 1 , 1), (X, x0 )]. In general, the fundamental group of a topological space depends upon the choice of basepoint ...
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... Let X and E be topological spaces and suppose there is a surjective continuous map p : E → X which satisfies the following condition: for each x ∈ X, there is an open neighborhood U of x such that • p−1 (U ) is a disjoint union of open sets Ei ⊂ E, and • each Ei is mapped homeomorphically onto U via ...
... Let X and E be topological spaces and suppose there is a surjective continuous map p : E → X which satisfies the following condition: for each x ∈ X, there is an open neighborhood U of x such that • p−1 (U ) is a disjoint union of open sets Ei ⊂ E, and • each Ei is mapped homeomorphically onto U via ...
Ab-initio construction of some crystalline 3D Euclidean networks
... H2 . The mutation rule is simple: the angle between intersecting mirrors of the orbifold must be divided by (b + 1), where b is the order of the corresponding point on the IPMS (zero for points on negative Gaussian curvature, positive for flat points) [10]. The corresponding Conway symbol entries fo ...
... H2 . The mutation rule is simple: the angle between intersecting mirrors of the orbifold must be divided by (b + 1), where b is the order of the corresponding point on the IPMS (zero for points on negative Gaussian curvature, positive for flat points) [10]. The corresponding Conway symbol entries fo ...
Solution 3
... Recall from lecture that for any topological space X with strongly discontinuous action of any group G, the map p : X → X/G is open and a covering map. Furthermore, if X is simply connected, π1 (X/G) ∼ = G. (a) To show that M/G is again a manifold, we check that it is second countable, locally Eucli ...
... Recall from lecture that for any topological space X with strongly discontinuous action of any group G, the map p : X → X/G is open and a covering map. Furthermore, if X is simply connected, π1 (X/G) ∼ = G. (a) To show that M/G is again a manifold, we check that it is second countable, locally Eucli ...