Let's Cover The Fundamental Group (pt 1)

Jumping right in


Firstly, this post builds off of my previous post, so if you’re learning these things for the first time and find yourself kinda lost, start there.

In addition to the previous post, the following are mathematical definitions you should probably know before reading this post. If you don’t know any of these, trust google.

Moving onward


Given a path $\alpha$ from $x_0$ to $x_1$ in $X$, we define $\hat\alpha:\pi_1(X,x_0)\to\pi_1(X,x_1)$ as follows:

And it’s a theorem that $\hat\alpha$ is a group isomorphism.


Given a continuous map $h:(X,x_0)\to(Y,y_0)$, we define:

$h_\star:\pi_1(X,x_0)\to\pi_1(Y,y_0)$ as follows:

$h_\star([f]) = [h\circ f]$.

And with that…

Let’s get into the questions!


Show that if $A\subseteq\mathbb{R}^n$ is star convex then $A$ is simply connected.

Let $a\in A$ be one of the points that make the set star convex. Our task now is to show:

  1. For any $x,y\in A$, there is a path entirely within $A$ connecting $x$ and $y$, and
  2. $\pi_1(A,a)$ is trivial.

For the first: just let $x,y\in A$ and let $f_x,f_y$ be the paths connecting them to $a$. Then note that $f_x \star \bar f_y$ is a path connecting $x$ to $y$.

For the second we can use the same straight-line homotopy that we used for the convex version: let $f$ be a path starting and ending at $a$ and let $H(s,t) = t\cdot a + (1-t)\cdot f(s)$.

Lastly, because part a (that I didn’t mention) is to find a set that’s star convex but not convex, I’ll leave the question with The Star of David.

Show that if $\alpha,\beta$ are paths from $x_0\to x_1\to x_2$ all in $X$ and $\gamma = \alpha\star\beta$ that $\hat\gamma = \hat\beta\circ\hat\alpha$.

Since these are functions between equivalence classes, our job is now to show that the outputs for a given input are homotopic.

So let $f$ be a path in $X$ starting and stopping at $x_0$. Then:

Show that if $x_0,x_1$ are points in a path-connected space $X$, $\pi_1(X,x_0)$ is abelian if and only if for every pair of paths from $\alpha,\beta$ from $x_0$ to $x_1$, $\hat\alpha = \hat\beta$.

If $\pi_1(X,x_0)$ is abelian, we have:

Let $A\subset X$ and let $r:X\to A$ be a retraction. Show that for $a_0\in A$, $r_\star:\pi_1(X,a_0)\to\pi_1(A,a_0)$ is surjective.

Well, any path $\alpha$ in $A$ starting and stopping at $a_0$ will also be a path in $X$ (because $A\subset X$). So $r_\star([\alpha]_X) = [\alpha]_A$. It’s surjective because it’s the identity map when we restrict paths to $A$.

Let $A\subset\mathbb{R}^n$ and $h:(A,a_0)\to(Y,y_0)$. Show that if $h$ is extendable to a continuous map $\tilde h:\mathbb{R}^n\to Y$, then $h_\star$ is trivial (i.e. sends everybody to the class of the constant loop).

Let $G=\pi_1(A,a_0), H=\pi_1(Y,y_0)$. Then, as a reminder, $h_\star:G\to H$ such that $h_\star([\alpha]) = [h\circ\alpha]$.

So let $\alpha,\beta$ be paths in $(A,a_0)$. Since $\alpha,\beta$ were arbitrary, it is sufficient to show that $h_\star([\alpha])$ is homotopic to $h_\star([\beta])$.

Consider $F:I\times I\to \mathbb{R}^n$ given by $F(s,t) = t\alpha(s) + (1-t)\beta(s)$ (the straight line homotopy between the two loops). But since $h$ is extendible to $\tilde h:\mathbb{R}^n\to Y$, we know that even $F$ is a homotopy that leaves $A$ (into some other part of $\mathbb{R}^n$), $\tilde h\circ F$ is a homotopy between $h(\alpha),h(\beta)$ that stays entirely in $Y$. Hence $[\tilde h\circ \alpha] = [\tilde h\circ \beta]$.