A question about the proof of the fact that contractible spaces are simply connected

Question

In greeberg's algebraic topology, the following fact is used in the proof that contractible spaces are simply connected without justification:

Let $p:\mathbb{I}\rightarrow X$ be a continuous function such that $p(0)=p(1)$ and $X$ is contractible. Then there exists a continuous function $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ such that:

1) $F(s,0)=p(0)$ for all $s\in \mathbb{I}$

2) $F(s,1)=p(s)$ for all $s\in \mathbb{I}$

3) $F(0,t)=F(1,t)$ for all $t\in \mathbb{I}$

I agree that there exists a map $F$ such that the first 2 conditions are satisfied, but what about the third ? Is there an obvious way to see that that the three conditions can be satisfied. As I mentioned the book uses this fact without justification.

Thank you

Reminder: All answers posted so far are proving the fact that contractible spaces are simply connected. Note that my original question (found above) asks for a justification of the fact that the book stated without justification. I don't think it s obvious.

Answer 1

A space $X$ is simply connected if and only if every continuous map $p: S^1\to X$ extends to a continuous map $P: D^2\to X$. Now, if you have a contraction $F: X\times I\to X$ of $X$, then simply take $$ P(r e^{it})= F(p(e^{it}), r). $$ Here I am regarding $D^2$ as the unit disk in the complex plane.

Answer 2

I apologize for the previous false starts, here's a simple way to use contractibility to construct an $F$ satisfying (1), (2) and (3). Notice, in particular, that such an $F$ does not need to have $F(0,t)$ constant, just needs to have it equal to $F(1,t)$ for each $t$.

OK, let $H : X \times [0,1] \to X$ be a homotopy of the identity on $X$ to the constant map with value $p(0)$, that is, let $H(x,0) = p(0)$, $H(x,1) = x$, for all $x$ (note that we do not assume that $H$ fixes $p(0)$ for all $t$, as that cannot always be arranged). Now define $F(s,t) = H(p(s),t)$. We have that

• $F(s,0) = H(p(s),0) = p(0)$,
• $F(s,1) = H(p(s),1) = p(s)$, and
• $F(0,t) = H(p(0),t) = H(p(1),t) = F(1,t)$.

Answer 3

I think the question is asking about proving that for a contractible space $X$ a map $(S^1,1) \to (X,x)$ is null homotopic relative to the base point $1$. Of course a contracting homotopy of $X$ may move the base point $x$.

It is often easier to understand these matters from a more general viewpoint. The key property of $(S^1,1)$ is that it is well pointed, i.e. has the homotopy extension property. So a relevant lemma is essentially as follows, and is 7.2.11 of Topology and Groupoids. Recall an inessential map is one homotopic to a constant map.

Let $f:Y \to X$ be an inessential map, and suppose $y$ is well pointed in $Y$. Then $f$ is inessential rel $y$.

The feature of the proof is that that the null homotopy of $f$ defines a path $\alpha$ in $X$ from $x=f(y)$ to $x'$ say. Because $(Y,y)$ is well pointed, $\alpha$ defines a bijection $\alpha_*: [(Y,y), (X,x)] \to [(Y,y), (X,x')]$, where these are homotopy classes relative to the base point. Let $g: Y \to X$ be the constant map with value $x$. Then $\alpha_*[f]= \alpha_*[g]$ and so $[f]=[g]$.

These operations generalise the operations of the fundamental group on higher homotopy groups. In moving from $(S^n,1)$ to $(X,A)$ and looking at the fact that a homotopy equivalence of spaces induces an isomorphism of homotopy groups, I discovered a gluing theorem for homotopy equivalences (Section 7.4 of the above book.)