We have the theorem which says that the induced homomorphism $p_* : \pi_1(\tilde X,\tilde x_0)\rightarrow \pi_1(X,x_0)$ is injective (hence a monomorphism). Here $\tilde X$ is a covering space of $X$.

I am just trying to digest this fact (the injectivity), since we know many spaces (e.g. wedge of two circles $S^1 \vee S^1$) have covers whose fundamental group is free group on many (or even infinite) generators.

My understanding of Injectivity (as it relates to groups) is that it implies that $\pi_1(X)$ is at least as "big" as $\pi_1(\tilde X)$, so I am confused here.

Confusion: The $\pi_1(X=S^1\vee S^1)$ is free group on two generators. $\pi_1(\tilde X)$ can be free group on many generators. How is the induced map injective then ?

I am not very experienced with group theory, though I have read upon the basics before reading Hatcher/Munkres for Algebraic Topology.

  • 1
    $\begingroup$ I cannot see what your question is. $\endgroup$ Commented Jun 17, 2013 at 22:00
  • $\begingroup$ Edited to add more info, thanks. $\endgroup$ Commented Jun 17, 2013 at 22:05

1 Answer 1


The free group on two generators contains subgroups which are free on as many generators as you want (and one particularly nice way of proving this is noticing that $S^1\vee S^1$ has finite wedges of circles as covering spaces :-) )

Indeed, if $L$ is a free group on two generators, one can check that the derived subgroup $[L,L]$ is in fact free on countably infinite generators!

  • 2
    $\begingroup$ and the lesson here is that free groups are weird. there's a natural instinct to treat them simply as the nonabelian analogues of free abelian groups, but the first thing we think of (rank) really doesn't translate. another thing that comes out of the (seemingly) innocuous $F_2$ is the banach-tarski paradox. it's trivial to see that it acts paradoxically, and fairly intuitive that almost all pairs in $SO(3)$ generate a free group. from that, it's just a small step to getting a paradoxical action on the unit ball... $\endgroup$ Commented Jun 17, 2013 at 22:19
  • $\begingroup$ (of course, one should also note that the banach-tarski paradox really isn't saying something that can be translated into cutting oranges. at least not with any knife i've seen) $\endgroup$ Commented Jun 17, 2013 at 22:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .