Show that the upper half plane is a simply connected region 
Let $H\subset \mathbb{C}$ be the upper half plane. Show that $H$ is simply connected.

From the definition, I want to show that for any two curves in $H$ with the same endpoint are homotopic. Also I know that $H$ is convex, so if $\alpha, \beta\in H$ then the line segment joining $\alpha$ and $\beta$ are also in $H$, denoted by $l$. Suppose $\sigma_1, \sigma_2$ are two curves with endpoints $\alpha,\beta$. We can continuously transform $\sigma_1$ to $l$ by $(1-s)\sigma_1+sl$. Similarly, we can continuously transform $l$ to $\sigma_2$. Can I concatenate the two transformations and say that we can continuously transform $\sigma_1$ to $\sigma_2$? Is my idea correct?
 A: Another way to see it is to observe that convex sets are contractible, and contractible spaces are simply connected.
The slickest proof of the first implication is that if $X$ is convex, then fixing any point $x \in X$, all other points in $X$ may be connected to $X$ by a straight line. Contracting along this line gives the desired homotopy. Now for the second implication, note that the fundamental group is a homotopy invariant.
A: Don't show two curves are homotophic, show any curve is homotopic to a single point and then from there you can use the fact that homotopy is an equivalence relation to get that any two curves are homotopic by composing $H_1:\gamma_1\times [0,1]\to *$ and $H_2^{-1}: *\times [0,1]\to \gamma_2$.
To show any curve is homotopic to a point, just use the straight line homotopy, from $\gamma\to *$, i.e draw the line from $\gamma(t), 0\le t\le 1$ to $*$ and progress from $\gamma(t)$ to $*$ at unit speed along the line. (Note this also shows that all convex sets are simply connected at the same time!)
A: A region $D$ is simply connected if both $D$ and $[\mathbb C\cup\{\infty\}]\setminus D$ be connected. But $H$ and $[\mathbb C\cup\{\infty\}]\setminus H$ are convex and then connected.
