Proof that every path in U is homotopic

Question

Let $U$ a connected space and given $x,y,x',y'$ four points $\in U $ and it says that every statement implies the other :

1) Every path ( in $U$ ) from $x \longrightarrow y $ is homotopic in $U$

2) Every path ( in $U$ ) from $x' \longrightarrow y' $ is homotopic in $U$

3) Every loop ( in $U$ ) is homotopic in $U$

Here is what I did , Two continuous paths $\gamma_{1}$ and $\gamma_{2} $ are path-homotopic if :

$\forall t \in [0,1]\quad f(t,0) = \gamma_0(t)\text{ and } f(t,1) = \gamma_1(t)$ ; ( $f$ continuous function defined for $[0,1]$)

in other words :

$\gamma_0(0) = \gamma_1(0) = x \text { and } \gamma_0(1) = \gamma_1(1) = y$

but it didn't prove anything , the only thing I know for sure is if the three conditions are true then $U$ is simply connected .

Any help would be a lot appreciated. Thanks in advance

Answer 1

First of all under the assumption that $U$ is only connected these are not equivalent.

Consider $U=T\cup S\subseteq\mathbb{R}^2$ where $T=\{(x,\sin(1/x))\ |\ x>0\}$ is the topologists's sine curve and $S=\{v\in\mathbb{R}^2\ |\ \lVert (-1,0)-v\rVert=1\}$ is the sphere around $(-1,0)$ of radius $1$. This space $U$ is connected, but not path connected. Actually $S$ and $T$ are the only two path components of $U$. Now if I choose $x,y\in T$ then every path $x\to y$ is homotopic, because $T$ is contractible. On the other hand not every path $x'\to y'$ is homotopic when $x',y'\in S$. And so conditions (1) and (2) can be true or false depending on the choice of $x,y,x',y'$. Note that $x,y,x',y'$ here are fixed, chosen before those conditions are stated. Otherwise conditions (1) and (2) would be the same.

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However these conditions are equivalent if $U$ satisfies stronger condition: being path connected.

With that we can generalize all of those conditions. First some notation. For $x,y\in U$ let $P(x,y)$ be the set of all paths from $x$ to $y$. Let $H(x,y)=P(x,y)/\sim$ be the quotient under "being homotopic" relationship. Then we have:

Lemma. If $U$ is path connected then for any $x,y,x',y'$ there exists a bijection $H(x,y)\to H(x',y')$. In particular all paths $x\to y$ are homotopic if and only if all paths $x'\to y'$ are homotopic.

Proof. Since $U$ is path connected then let $\lambda,\beta:I\to U$ be paths such that $\lambda(0)=x$, $\lambda(1)=x'$, $\beta(0)=y$, $\beta(1)=y'$.

Recall that if $f:I\to U$ is a path then $f^{-1}:I\to U$ is the path defined by $f^{-1}(t)=f(1-t)$. And so it is the path in the other direction.

Also if $f,g:I\to U$ are two paths such that $f(1)=g(0)$ then by $f*g$ I will denote the path composition of $f$ and $g$.

Now define

$$P:P(x,y)\to P(x',y')$$ $$P(\alpha)=\lambda^{-1}*\alpha*\beta$$

I leave as an exercise that $P$ is a well defined function. Note that formally there should be brackets here, because "$*$" is not associative. However it is associative under homotopy and so it doesn't really matter. For full correctness you can put brackets however you want, e.g. $\lambda^{-1}*(\alpha*\beta)$.

Unfortunately $P$ is almost never a bijection. However it does induce a bijection on homotopy classes. How do we show that? Well, the inverse is induced by $G(\delta)=\lambda*\delta*\beta^{-1}$ and so all we need to show is that there exists a homotopy

$$\lambda*\lambda^{-1}*\alpha *\beta*\beta^{-1}\simeq\alpha$$

And this is a consequence of the following three facts:

1. "$*$" is associative under homotopy, i.e. $f*(g*h)\simeq (f*g)*h$
2. $f*f^{-1}\simeq c$ and $f^{-1}*f\simeq c'$ where $c,c'$ are constant maps at appropriate points
3. if $c,c'$ are constant maps (at appropriate points) then $c*f\simeq f$ and $f*c'\simeq f$

Can you complete the proof?