Undergrad Topology - 400 Level
1. Prove that v(Г) - e(Г) = 1 for any tree T. (v :vertices and e : edges) 2. Even better, show that v(Г) - e(Г) ≤ 1 for any graph Г, with equality precisely when Г is a tree.
Find a tree in the polyhedron of figure 1.3 which contains all the vertices. Construct the dual graph Г and show that Г contains loops. (You don't have to construct the graph, but please describe it to me how it looks like.) (SEE ATTACHMENT)
14. Make a Mą¶bius strip out of a rectangle of paper and cut it along its central circle. What is the result? 15. Cut a Mą¶bius strip along the circle which lies halfway between the boundary of the strip and the central circle. Do the same for the circle which lies one-third of the way in from the boundary. What are the resul ...continues
Define f: [0, 1) → C by f (x) = e2πix. Prove that f is one-one, onto, and continuous. Find a point x ∈ [0, 1) and a neighborhood N of x in [0, 1) such that f (N) is not a neighborhood of f (x) in C. Deduce that f is not a homomorphism.
19. Let X be a topological space and let Y be a subset of X. Check that the so-called subspace topology is indeed a topology of Y. (question is also included in attachment)
Circles that Cannot be Homomorphic
23. Using the intuitive notion of connectedness, argue that a circle and a circle with a spike attached cannot be homomorphic. (Question is also included in attachment)
Annulus : X and Y cannot be Homomorphic
24. Let X,Y be the subspace of the plane shown as below. Under the assumption that any homomorphism from the annulus to itself must send the points of the two boundary circles among themselves, argue that X and Y cannot be homomorphic. (Question is also included in attachment)
Make a model for a Klein bottle
27. Make a model for a Klein bottle as shown below.... see attachment
Let X be a topological space and let Y be a subset of X. Check that the so-called subspace topology is indeed a topology on Y.
In a recent survey of 1282 moviegoers, 712 said they prefer butter on their popcorn while the rest did not. Now given the chance that at least one person will have a cell phone conversation right at the most intense part of the movie and given the fact that you have a 1 in 3 chance of sitting in front of a crying baby all ...continues
