# sol-03 - MATH 453 SOLUTIONS TO ASSIGNMENT 3 Exercise 1 from...

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M  453 S   A  3 S  26, 2004 Exercise 1 from Sections 14-16, page 91 First consider A as a subspace of Y . In this topology, a subset B of A is open if and only if it is of the form B = A V , where V is an open set in Y . Now Y is a subspace of X , so V is open precisely when it is of the form V = Y U for some U open in X . Putting these together, we get B = A ( Y U ) = A U , since A Y . Hence we reach the conclusion that B is open if and only if B = A U for some open set U of X , which is the very definition for B to be open in the subspace topology A inherits from X . square Exercise 4 from Sections 14-16, page 92 We’ll check that π 1 : X × Y X is an open map; the argument for π 2 is exactly the same. Let W be an open set in X × Y . Using the definition of the product topology, we can write W as a union of basis elements: W = uniondisplay α J U α × V α , where J is some indexing set, U α is open in X , and V α is open in Y . Unions behave well under a function, namely f ( uniontext α A α ) = uniontext α f ( A α ), and π 1 ( U α × V α ) = U α , so π 1 ( W ) = π 1 uniondisplay

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