Course 212 (Topology), Academic Year 1989—90
... Definition A topological space X is said to be connected if and only if the empty set ∅ and the whole space X are the only subsets of X that are both open and closed. Proposition 5.1 will provide some alternative characterizations of the concept of connectedness. First we make some observations conc ...
... Definition A topological space X is said to be connected if and only if the empty set ∅ and the whole space X are the only subsets of X that are both open and closed. Proposition 5.1 will provide some alternative characterizations of the concept of connectedness. First we make some observations conc ...
PDF
... manifolds such that the group operations are smooth maps. Other groups • The trivial group consists only of its identity element. • The Klein 4-group is a non-cyclic abelian group with four elements. For other small groups, see groups of small order. • If X is a topological space and x is a point of ...
... manifolds such that the group operations are smooth maps. Other groups • The trivial group consists only of its identity element. • The Klein 4-group is a non-cyclic abelian group with four elements. For other small groups, see groups of small order. • If X is a topological space and x is a point of ...
the fundamental group and covering spaces
... The subset p∗ π(Y )(y1 , y2 ) ⊂ π(X)(x1 , x2 ) consists of all paths from x1 to x2 that lift to paths from y1 to y2 . The subbgroup p∗ π1 (Y, y0 ) ≤ π1 (X, x0 ) consists of all loops at x0 that lloft to loops at y0 . Definition 2.8. The monodromy functor of the covering map p : X → Y is a functor F ...
... The subset p∗ π(Y )(y1 , y2 ) ⊂ π(X)(x1 , x2 ) consists of all paths from x1 to x2 that lift to paths from y1 to y2 . The subbgroup p∗ π1 (Y, y0 ) ≤ π1 (X, x0 ) consists of all loops at x0 that lloft to loops at y0 . Definition 2.8. The monodromy functor of the covering map p : X → Y is a functor F ...
Math 490 Extra Handout on the product topology and the box
... 6. Define f : R → RN by setting f (t) = (t, t, t, t, . . . , t, . . .) for all t ∈ R. Prove that f is not continuous if we give RN the box topology. Notice that this implies that the result of the previous exercise fails for the box topology. 7. Show that RN is disconnected in the box topology. (Hi ...
... 6. Define f : R → RN by setting f (t) = (t, t, t, t, . . . , t, . . .) for all t ∈ R. Prove that f is not continuous if we give RN the box topology. Notice that this implies that the result of the previous exercise fails for the box topology. 7. Show that RN is disconnected in the box topology. (Hi ...
