For each of the following subgroups of

$$ \left \langle x,y \right \rangle = \pi_{1}(S^{1}\vee S^{1}) $$

construct a based covering map

$$ \ p:(\tilde{X},\tilde{b})\rightarrow (S^{1}\vee S^{1},b) $$

such that

$$ p_{\ast }\pi_{1}(\tilde{X},\tilde{b})\subset \pi_1(S^1\vee S^1,b) $$

is that subgroup:

i) $$ \left \langle x \right \rangle $$

ii) $$ \left \langle x^{n_1}y^{m_1}...x^{n_k}y^{m_k} : \sum{m}_i\ is\ even \right \rangle $$


the kernel of the homomorphism $$ \left \langle x,y \right \rangle\rightarrow \mathbb{Z}\times \mathbb{Z} $$

that sends x to (1, 0), and y to (0, 1).

Hi, I'm an undergrad doing algebraic topology and I'm really stuck on this homework question, I'm not very sure how to go about constructing the covering maps here at all, could someone give me a hint or a walk through of one of the questions please? Thanks.


You have to consider the action of $\pi_1(X,b)$ by deck transformations on the universal covering space $\widetilde{X}$. Then you take your given subgroup $\Gamma\subset \pi_1(X,b)$ and look at the quotient $\overline{X}=\Gamma\backslash \widetilde{X}$. It will turn out that $\pi_1(\overline{X},\overline{b})$ is isomorphic to $\Gamma$ and (because $p_*$ is injective) the same is true for its image under $p_*$.

In the case of $X=S^1\vee S^1$, the universal covering space is a 4-valent tree and the generators $x$ and $y$ act by "translations".

So for example the quotient by $\langle x\rangle$ looks like a copy of ${\mathbb R}$ with one circle attached to each integer point. You find more examples at pages 57-59 of http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf


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