If we delete a point from y we can get either 2 or 3

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we will get two connected pieces. If we delete a point from Y we can get either 2 or 3 connected pieces. Therefore X can not be topologically equivalent to any of N , Z , or Y . By the same reasoning Y can not be topologically equaivalent to N or Z . ±inally N and Z are clearly equivalent by distortion sunce we can distor each of them into a straight segment. The correct answer is (c) . Question 6. Which of the following objects is topologically equivalent to a cylinder with one hole:
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(a) a sphere with three holes (b) a torus with a hole (c) a hemisphere (d) a hemisphere with a hole (e) a cylinder (f) a sphere Answer 6. A cylinder has two boundaries - the two rims of the cylinder. If we cut an extra hole in the surface of the cylinder we will get another boundary. Out of the six listed surfaces only the sphere with three holes has three boundaries. The correct choice is (a) . Question 7. The edges of a rectangular sheet of fabric with a circular hole in the middle are identiFed as shown on the following picture: How many boundaries does the resulting surface have? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 (f) 5 Answer 7. If we identify the edges of a plain rectangular sheet in the indicated manner we will get a Klein bottle which has no boundaries. The extra hole we have cut out creates an additional boundary so we have a surface with one boundary. The correct answer is (b) . Question 8. Consider the following graph:
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What is the minimal number of edges we must add to this graph so that it contains an Euler circuit? Draw these edges clearly on the picture. (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 (f) 5 Answer 8. The graph has 6 vertices of valence 3, one vertex of valence 2 and one vertex of valence 4. To have an Euler circuit it is necessary and suFcient to have all vertices be of even valence. So we do not need to do anything to the vertices of valence 2 and 4. So, to guarantee that
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If we delete a point from Y we can get either 2 or 3...

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