2
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ A trivial group contains exactly one element, the identity $\endgroup$ – Alessandro Codenotti Jan 17 at 18:50
  • 2
    $\begingroup$ Your statement "$\alpha \simeq e_x$ and $\beta\simeq e_x$" means that all three loops are in the same homotopy class. If $\alpha \simeq e_x$ for all $\alpha$, then there is only one homotopy class of loops and it is $[e_x]$. $\endgroup$ – William Jan 17 at 21:52
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @ Chris Custer I guess my confusion arises from the fact that the fundamental group is defined as the set of homotopy equivalence classes of all loops in $X$. Hence, even though $\alpha$ and $\beta$ are null homotopic in the example I gave above, shouldn’t the fundamental group contain the equivalence classes of $\alpha$ and $\beta$ along with $e_x$, and hence be “non-trivial”? This would seem to contradict the claim. $\endgroup$ – JG123 Jan 17 at 21:35
  • 2
    $\begingroup$ But $[\alpha]=[\beta]=[e_x]$. $\endgroup$ – Chris Custer Jan 17 at 21:37
  • $\begingroup$ @ Chris Custer I came to the same conclusion. Hence, since all of the equivalence classes are the same, the fundamental group is trivial. Thank you so much for the help. $\endgroup$ – JG123 Jan 17 at 22:59
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.