Fibration $p_1:P(Y,y_0)\to Y$ has section iff $Y$ is contractible.

Question

Let $P(Y,y_0) = \{ \omega : \omega(0) = y_0 \}$ be path space let's consider a fibration $p_1:P(Y,y_0)\to Y$ such that $\omega \mapsto \omega(1)$. Show that there exists $s: Y \to P(Y,y_0)$ such that $p_1 \circ s = id_Y$ if and only if $Y$ is contractible space.

If $Y$ is contractible then clearly $s(y) = H(y, \square)$ (where $H$ is homotopy between $c_{y_0}$ and $id_Y$) fulfills assumptions. But I'm having a hard time with the remaining implication. The very first shot is obviously to put again $H(y,t):= s(y)(t)$ but it doesn't have to be continuous with respect to $Y \times I$ (we only know that it is continuous w.r. to $Y$ by continuity of $s$). My other tries with usage of fibration properties are failing at the same point - proving that my potential homotopy is continuous. Any ideas?

Somehow linked because the point may be my misunderstanding of what does continuity of $s$ really mean: Topology on the space of paths

Answer 1

As you said, let $H : Y \times I \to Y$ be defined by $H(y,t) = s(y)(t)$. Let's prove that it is continuous, then it will be the required homotopy. I'll do it with all the details, hopefully it will be clear enough.

Let $U \subset Y$ be open; we want to prove that $$H^{-1}(U) = \{ (y, t) \in Y \times I : s(y)(t) \in U \}$$ is open.

The space $P(Y, y_0) \subset C(I,Y)$ has the subspace topology induced by the compact-open topology. We know the map $s$ is continuous, which means that for all subsets $$B(K,V) = \{ \omega \in P(Y,y_0) : \omega(K) \subset V \}$$ for $K \subset I$ compact and $V \subset Y$ open, the inverse image $$s^{-1}(B(K,V)) = \{ y \in Y : \forall t \in K, \; s(y)(t) \in V \}$$ is open.

Now let's take $(y,t) \in H^{-1}(U)$ and look for a neighborhood entirely contained in $H^{-1}(U)$. We know $s(y)(t) \in U$; since $s(y)$ is continuous, there's a small neighborhood $t \in J \subset I$ such that $s(y)(J) \subset U$. Take a closed interval (thus compact) $K \subset J$ such that $s(t)$ is in the interior of $K$ (that's always possible).

Then we know that $s^{-1}(B(K,U)) \times \operatorname{int}(K) \subset Y \times I$ is open (it's the product of two open sets), it is entirely contained in $H^{-1}(U)$ (check the definitions), and it contains $(y,t)$. That's what we wanted.

Answer 2

I realized that your question was more about the compact-open topology than about the problem itself. Thereby, the answer given by Najib Idrissi was very appropriate. However, let me give you an alternative answer. Firstly, you have to observe that the contractible spaces are final in the homotopy category $ Ho(Top) $ of topological spaces and homotopy classes of continuous functios between them.

Secondly, you have to observe that $P(Y, y_0) $ is contractible (and, ideed, $P(Y, y_0) $ may be seen as the fibrant replacement of the point). Actually, here you have to understand the compact-open topology!

Then (and this is "abstract nonsense") every morphism $P(Y, y_0)\to Z $ in $Ho(Top) $ is a section of the unique morphism $ Z\to P(Y, y_0) $ in $ Ho (Top ) $. In particular, if you have such a morphism $ [R]: Z\to P(Y, y_0 ) $ which is also a retraction in $Ho(Top)$ (as in your case), then $ [R] $ is a homotopy equivalence.

In other words, you have just to realize that $X$ is final in $Ho(Top)$ if and only if $X$ is contractible. And, then, you have to use the fact that retractions of final objects are final objects!

On the other hand, If you prefer avoiding categories, you have just to observe that, since $P(Y, y_0)$ is contractible, if X is any topological space, then there is an unique homotopy class of continuous functions $X\to P(Y,y_0) $. And, because of that, if $p: P(Y, y_0)\to X$ is a continuous function, you may compose with any continuous function $s: X\to P(Y, y_0) $ and get a function

$s\circ p: P(Y, y_0)\to P(Y, y_0) $

which is homotopic to the identity. So, if you assume by hypothesis that $p\circ s $ is identity, you proved that $X$ has the same homotopy type of $P(Y,y_0)$. I.e. $X$ is contractible!

The general theorem is the following: any retraction (retract) of a contractible space is contractible. So, you don't need to use the fact that $P(Y,y_0)\to Y $ is a fibration to give a solution to your problem!