# starting with paths and the fundamental group

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.

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## 1 Answer

For (i) you've done a lot of the work, but some care is needed. You have indeed shown that any $f\colon X \to Y$ is homotopic to a constant map, but if $Y$ is not path connected then there may be different constant maps from $X$ to $Y$ that are not homotopic. As a silly example, consider the space of maps $\{x\} \to \{y_1, y_2\}$, where these finite sets have the discrete topology.

So, picking up from where you leave off and using that homotopy is an equivalence relation, suppose we have constant maps $f, g\colon X \to Y$. As $Y$ is path connected, there is a path $\gamma$ from $g(k_0)$ to $h(k_0)$. Use this to construct a homotopy between $f$ and $g$ (At this stage, $X$ does not play a significant role).

I would also avoid thinking of these maps $X \to Y$ as paths. A path in $Y$ is a map $S^1 \to Y$, and $S^1$ is not nullhomotopic. So you certainly can't conclude that $Y$ is simply connected, and indeed that need not be the case.

For (ii), I think you have the right idea; you just need to make an argument. Take an open neighborhood $U$ of $b$ such that $p^{-1}(U)$ is a disjoint union of $\{V_i\}_{i \in I}$, each $V_i$ mapping homeomorphically onto $U$. If $b'$ is another element of $U$, then define a map $\Phi\colon p^{-1}(b) \to p^{-1}(b')$ as follows. If $e \in p^{-1}(b)$, then $e$ is an element of $V_i$ for some (unique) $i \in I$. Show that there is a unique element $e'$ of $V_i$ such that $p(e') = b'$. Set $\Phi(e) = e'$, and show that this is a bijection.

It follows that if $p^{-1}(b)$ has $k$ elements, then so does $p^{-1}(b')$ for $b'$ in an open neighborhood around $b$. You can say something similar if $p^{-1}(b)$ does not have $k$ elements. Now use connectedness.

Let me know if I should say more!

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And How can I use the hypothesis of being path connected in i) ? – Daniel Oct 30 '11 at 18:28
@Daniel Updated. – Dylan Moreland Oct 30 '11 at 18:48
Thanks! And in the problem ii) At least the problem, it´s from Munkres, section 53, problem 3. I don´t know If I need more hypothesis or not. – Daniel Oct 30 '11 at 19:22
@Daniel Woops, paths on the brain, I guess. Fixed it up. – Dylan Moreland Oct 30 '11 at 19:45
I´m not sure, how to construct, the new homotopy D: – Daniel Nov 1 '11 at 20:19
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