I have two problems, I´m a little complicated with the problem , I know that it´s easy but I need just a little help. Are two problems
i) Suppose that the identity map $ i:X \to X $ is homotopic to a constant path, and the space Y is path connected, then the set of homotopy classes of maps of X into Y has a single element.
ii) Let $ p:E \to B $ be a open covering , B connected. Show that if for some $ b_0 $ we have $ \left| {p^{ - 1} \left( b_0 \right)} \right| = k\,\,\, $ then for every b we have $ \left| {p^{ - 1} \left( b \right)} \right| = k $
My try:
i) Im not sure if this is ok, but i "proved" that any path $ f:X \to Y $ is homotopic to a constant map, and this implies that Y is simply connected. But i did not used the fact that Y is path connected, I think that something that I did, is wrong. Let f be a path from X to Y. Since the identity in X is nulhomotopic, then there exist a homotopy $ F:I\times X \to X $ such that: $ \eqalign{ & F\left( {0,x} \right) = i\left( x \right) = x \cr & F\left( {1,x} \right) = k_0 \in X \cr} $
Then the homotopy $ H = f \circ F:I\times X \to \,Y $ "deforms f to a point " since $ \eqalign{ & H\left( {0,x} \right) = f\left( x \right) \cr & H\left( {1,x} \right) = f\left( {k_0 } \right) \in Y \cr} $
For ii) I suppose that the set of elements such that its preimage is "k" (for every k ) elements must be open, but I´m not sure if this is true, if this is not true, I have no idea , How to do the problem.
Help please Dx