Let $(X,\ast)$ be a based topological space (maybe path connected or not, I don't know if this will be relevant to the solution).

Let $\pi:=\pi_1(X,\ast)$ be its fundamental group and let $H$ be any nontrivial subgroup of $\pi_1$ (suppose that there exists one).

My question is: is there some subspace $Y\subset X$ such that $H=\pi_1(Y,\ast)$?

In resume: given a subgroup of a fundamental group, is it also a fundamental group of some subspace of the total space?

  • $\begingroup$ How much do you know about coverings? $\endgroup$ – Neal Apr 30 '13 at 0:27
  • $\begingroup$ Some facts. It's hard to write. What are you thinking? Some hints? $\endgroup$ – Sigur Apr 30 '13 at 0:29
  • $\begingroup$ Just that if you're interested in realizing spaces whose fundamental groups are subgroups of $\pi_1(X)$, the natural place to look is covers, not subspaces. $\endgroup$ – Neal Apr 30 '13 at 2:58
  • $\begingroup$ @Neal, thanks. The question arose when I was studying other kinds of subgroups which have this kind of property. I'll think about. $\endgroup$ – Sigur Apr 30 '13 at 11:35

I'm interpreting your question as follows: when you write $H = \pi_1(Y,\ast)$, you mean the inclusion $i:Y\rightarrow X$ induces an injective map on $\pi_1$ with image $H$. (As opposed to $\pi_1(Y)$ being abstractly isomorphic to $H$).

With this interpretation, the answer is no. For example, consider $X = S^1$. It is well known that $\pi_1(X) = \mathbb{Z}$. Now, consider the subgroup $H = 2\mathbb{Z}$. I claim that there is no proper subspace $Y$ for which which has this fundamental group.

We may assume wlog that $Y$ is connected. Then note that any connected proper subset of $S^1$ is homeomorphic to a connected subset of $(0,1)$. These are easy to classify, they are, up to homeomorphism, $(0,1)$, $[0,1]$, and $(0,1]$. None of these has infinite cyclic fundamental group.

Edit Here is an example with $H$ not even abstractly isomorphic to a particular subgroup of $\pi_1(X)$.

Take $X = S^1 \vee S^1$, the wedge sum of 2 $S^1$s. The fundamental group of $X$ is known to be isomorphic to the free group on two generators. It's also know that a free group on two generators contains subgroups isomorphic to the free group on $n$ generators for any finite $n$ (we make even take $n$ to be countable). Further, it's know that free groups on different numbers of generators are never isomorphic.

Let $H$ denote any of these subgroups for $n > 2$. In particular, $H$ is not isomorphic to either $0$, $\mathbb{Z}$, or the free group on two generators. I claim that no subspace $Y$ of $X$ has $H$ as a fundamental group, even up to abstract isomorphism.

As above, we may assume $Y$ is connected. If $Y$ does not contain the wedge point, then it must be contained in a proper portion of one of the two circles, so the above argument shows $\pi_1(Y)$ is trivial. Hence, $Y$ must contain the wedge point. Now, if $Y$ does not contain the whole of the first circle, then $Y$ deformation retracts onto a subspace of the other circle, hence, by the previous argument has $\pi_1 = 0$ or $\mathbb{Z}$. Thus, $Y$ must contain the while first circle. Likewise, $Y$ must contain the whole second circle.

But then this implies $Y = X$, so $Y$s fundamental group is isomorphic to the free group on $2$ generators, so not isomorphic to $H$.

Final (?) Edit Here's one in the finite fundamental group case. Let $X$ be obtained from $S^1$ by attaching a $D^2$ by a degree $4$ map. A simple van Kampen argument shows $\pi_1(X) = \mathbb{Z}/4$. Let $H$ be the unique subgroup of $\pi_1(X)$ isomorphic to $\mathbb{Z}/2$.

I claim that no subspace $Y$ has fundamental group abstractly isomorphic to $H$. If $Y$ misses a point of the interior of the $D^2$, then $Y$ deformation retracts onto a subspace of $S^1$, so by the above argument, has $\pi_1 = 0$ or $\mathbb{Z}$. Hence, we may assume wlog that $Y$ contains all of the interior of $D^2$. Likewise, if $Y$ misses a point of $S^1$, then restricting the van Kampen argument to $Y$ shows that $\pi_1(Y) = 0$, so $Y$ must contain all of $S^1$. This implies $Y = X$, so $\pi_1(Y)\neq H$.

  • $\begingroup$ Great! What about both finite groups? $\endgroup$ – Sigur Apr 30 '13 at 0:02
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    $\begingroup$ Well, I answered the more general question, but with infinite groups. I'll think about the finite group case. My guess is that a similar argument would work, but I'm not seeing it. $\endgroup$ – Jason DeVito Apr 30 '13 at 0:04
  • $\begingroup$ I believe that you can delete the ? above.. lol $\endgroup$ – Sigur Apr 30 '13 at 0:52
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    $\begingroup$ @Sigur: I didn't mean anything negative by it, only that I may keep thinking about it. I like my counterexamples to be smooth, closed manifolds when they can be. So, there may be another edit in the future ;-). If you'd prefer, I can delete the "?". Thanks for asking such a fun question! $\endgroup$ – Jason DeVito Apr 30 '13 at 0:58
  • $\begingroup$ OK, no problem. The example is yours. Thanks so much for this. Sometimes, simple questions have elegant solutions. $\endgroup$ – Sigur Apr 30 '13 at 1:10

Take the space $X:=\mathbb{S}^1$ then $\pi_1(X)=\mathbb{Z}$ But any proper subspace of $X$ has has trivial fundamental group, so no subspace has fundamental group $\mathbb{2Z}\leq\mathbb{Z}$.


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