Definition of Simple Connectedness

Question

I was given the following definition for "simple connectedness":

"Let $X$ be an arcwise connected, locally arcwise connected space. Then, X is simply connected if its fundamental group is trivial, or equivalently, if every closed path in $X$ is homotopic to a constant."

I seek to prove this "equivalence" claim.

Let's assume that every closed path in $X$ is homotopic to a constant, $e_x$. (Recall that $e_x(t) = x$, where $x \in X$ and $t$ is on some interval). Let $\alpha$ and $\beta$ be homotopic to $e_x$. That is, $\alpha \simeq e_x$ and $\beta \simeq e_x$. Since homotopy is an equivalence relation, it follows that $\alpha \simeq \beta$. Therefore, $\alpha \in <\beta>$, where $<\beta>$ is the homotopy equivalance class of $\beta$. Hence, $<\beta>$ is nonempty. Now, the fundamental group of $X$ is the group of homotopy equivalence classes of loops in $X$. My question is this: How can the fundamental group be trivial, since it contains $<\beta>$?

Thanks in advance.

Answer 1

The fundamental group is trivial iff it has only one element iff the only homotopy class is the trivial one iff every loop is homotopic to the identity, which is the constant loop.

To try and answer your question, the idea is that, when the fundamental group is trivial, every loop $\beta$ is null homotopic.

So, while $\beta$ may not be the identity ( or constant loop), it is homotopic to it. For instance, this is true on the sphere, $S^2$, by an easily visualizable homotopy, or deformation.

What you're forgetting is that we're talking about homotopy classes of loops, not just loops themselves. This "trick" is sort of what makes the fundamental group a tractable object, in many cases.

Or are you forgetting that the trivial group is different than the empty set. The trivial group contains the identity, $e$, just as every group must.

Answer 2

Note that you've shown that $\alpha \cong e_x$ and $\beta \cong e_x$. You also noticed that homotopy is an equivalence relation. So not only can you conclude that $\alpha \in [\beta]$, but also that $\beta \in [e_x]$. So $\alpha \in [e_x]$. Since you picked $\alpha$ and $\beta$ arbitrarily, you have in fact shown that any loop is in the same homotopy class as the constant loop. But that's just saying that every loop is null homotopic, as required.