Simple Connectedness in the Plane

Question

A set is simple-connected if any two paths with the same endpoints are (path) homotopic. Show that this is equivalent to any loop being homotopic to the trivial loop.

Answer 1

First, let's be explicit: what you want to prove is:

A set is simply-connected if any two paths with the same endpoints are homotopic rel. $\{0,1\}$ (which is to say that the homotopy keeps the endpoints fixed). Show this this is equivalent to any loop being homotopic to the trivial loop rel. $\{0,1\}$.

Now, let me get the idea out and then we get our hands dirty.

An homotopy between two paths $c_0,c_1$ with the same endpoints (say, $x_0,x_1$) rel. to $\{0,1\}$ is a completion of a mapping $H$ defined on the boundary of the square $I \times I$ as $c_0$ on top, $c_1$ on bottom, $x_0$ on the left side and $x_1$ on the right side. Pictorically, we have such an homotopy if we can complete the square below.

So, to prove the "$\implies$" part, we must show that we can complete the square below for any loop $c$.

If we consider $c_1$ as the concatenation of itself up to half and itself after half (let's call them $c_1^{-},c_1^{+}$, respectively), we can investigate the square below.

We can obviously fill up the side triangles, by making $H$ to be constant on vertical lines (the $c_1^+$ and $c_1^-$ in the middle are obviously intended to be reparametrized accordingly to the increase of the length which they are now defined upon). So the question is if we can fill up the middle. However, our hypothesis gives us this, since filling up the middle is equivalent to finding an homotopy with endpoints fixed between $c_1^{-}$ and $c_1^{+}$ (this last one reversed).

The drawings are all fine, but how does this translate to practice?

Well, the condition that $H$ be constant on the side triangles can be stated as:

$$H(t,s):=c_1(t), \quad s \geq 2t \text{ or } s \geq 2-2t.$$

Note that this tells us that $H(t,1)=c_1(t)$ for all $t$, so everything is going fine. Now, we are supposed to use our homotopy between $c_1^{-}$ and $c_1^{+}$, call it $F$ (important: this homotopy is between $c_1^{-}$ and $c_1^{+}$ reparametrized accordingly to $[0,1]$! I could change notation, but this would be overload of notation), and suppose that $F(\cdot,1)=c_1^{-}$ and $F(\cdot, 0)=c_1^+$ . Therefore, the natural choice (it should be a good exercise to see why this is a natural choice) is to set $$H(t,s):=F(s,\frac{1}{1-s}(-t+1/2)+1/2), \quad s \leq 2t \text{ and } s \leq 2-2t. $$ There is, however, the problem $s=1$ above. If there were not this problem, the glueing lemma would show that $H$ is continuous on the square (for instance, as last defined, $$H(t,2t)=F(2t,\frac{1}{1-2t}(-t+1/2)+1/2)=F(2t,1/2+1/2)=F(2t,1)=c_1^{-}(2t),$$ as we have carefully concocted). It suffices then to show that the last piece of the function $H$ defined on the middle triangle as $$H(t,s):=F(s,\frac{1}{1-s}(-t+1/2)+1/2);$$ $$H(1/2,1)=x_0$$ is continuous at $(1/2,1)$. For this, take a neighbourhood $U$ of $x_0$. The pre-image under $F$ of $U$ is a neighbourhood of the right side of the square $I \times I$. By compactness of the right side of the square, this neighbourhood must have a tube of the form $V \times I$ inside it, where $V=(a,1]$ (this is essentially the tube lemma). Considering the function $$\phi: (t,s) \mapsto (s,\frac{1}{1-s}(-t+1/2)+1/2),$$ it is clear that if we take a neighbourhood $W$ of $(1/2,1)$ of "length" less than $1-a$, this function sends this neighbourhood inside $U$, because it sends it inside $V \times I$ (the function above is a clockwise rotation composed with a translation). Therefore, $F \circ \phi=H$ sends $W$ inside $U$, and continuity of $H$ on the triangle everywhere follows. This proves everything we wanted (just check that $H$ coincides with what it must on where it must).

Just to be quite explicit, we then have that, given a homotopy $F$ rel. $\{0,1\}$ between $c_1^{-}$ and $c_1^{+}$ (reversed), the function $$H:I \times I \to X$$ $$H(t,s):=c_1(t), \quad s \geq 2t \text{ or } s \geq 2-2t.$$ $$H(t,s):=F(s,\frac{1}{1-s}(-t+1/2)+1/2), \quad s \leq 2t \text{ and } s \leq 2-2t, s \neq 1 $$ $$H(1/2,1):=x_0$$ fits the bill.

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This is, however, only one implication. The other is quite similar, so I'll leave for you as an exercise. To simplify, I will put the diagram on which you should reason as we did above.

You should understand the following points:

• You want to fill the entire square (why?).
• It is trivial to fill the top and bottom triangles (why?).
• The hypothesis says you can fill the left triangle (why?).

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Final note: there usually are "smart" suggestions of homotopies which are simpler. However, the approach above is (although laborious), quite straightforward and geometric. In fact, after two or three examples, the procedure is quite clear, and it becomes insightless to repeat it, so most topologists simply argue based on the pictures (which is only justified, of course, if you know the procedure).

Answer 2

Let $\gamma_1:[0,1]\mapsto\mathbb{R}^2$ and $\gamma_2:[1,2]\mapsto\mathbb{R}^2$ be two paths with the same endpoints. Then we can define $\gamma:[0,2]\mapsto\mathbb{R}$ in the following manner

$$\gamma(t):=\cases{\gamma_1(t),\text{if $t\in[0,1]$} \\ \gamma_2(3-t),\text{if $t\in[1,2]$}\\}$$

So $\gamma$ is is a loop, because $\gamma(0)=\gamma_1(0)=\gamma_2(1)=\gamma(2)$ and we can conclude that $\gamma$ is homotopic to a point $x_0$. Let $\Gamma:[0,1]\times[0,2]\mapsto\mathbb{R}^2$ be such homotopy, so the following facts hold

$\cdot \ \Gamma(0,t)=\gamma(t)$

$\cdot \ \Gamma(1,t)=x_0$

Consider now the following funtion $\psi:[0,2]\times[0,1]\mapsto\mathbb{R}^2$:

$$\psi(s,t):=\cases{\Gamma(s,t),\text{ if $0\leq s\leq1$}\\ \Gamma(2-s,2-t),\text{ if $1\leq s\leq2$}}$$

And notice that $\psi$ is continuous and satisfy:

$\psi(0,t)=\gamma_1(t)$

$\psi(2,t)=\gamma_2(t+1)$

And, a linear transformation can turn this into an homotopy.

Obs: It took a little long for me to figure which linear funtions to use inside the brackets. Take little while to see why(or if) they are correct.