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Hi. I have just calculated the fundamental group of this pentagonal region but I am not sure if it is correct. I attach my solution.


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Let $U_1=X-\left\{y\right\},\, U_2=X-(a_1\cup a_2)$ open sets and path connected.

The figure 8 is a strong deformation retract of $U_1$ then $\pi_1(U_1,x_1)=\pi_1(\text{ Fig. 8 },x_1)=\langle [\alpha_1],\, [\alpha_2];\emptyset\rangle$ with $\alpha_1,\,\alpha_2$ closed path in $x_1$.

Therefore, $\pi_1(U_1,x_0)=\langle [\delta*\alpha_1*\overline{\delta}],\,[\delta*\alpha_2*\overline{\delta}];\emptyset\rangle=\langle A_1,A_2;\emptyset\rangle$ with $A_i=[\delta*\alpha_i*\overline{\delta}],\, i=1,2.$

Now, $\pi_1(U_2,x_0)=1$ because $U_2$ is contractible and $\pi_1(U_1\cap U_2,x_0)=\langle [\gamma];\emptyset\rangle=\mathbb{Z}$ because the circle generated by $\gamma$ is a strong deformation retract of $U_1\cap U_2$.

On the other hand, $\varphi_{1*}([\gamma])=[\delta*\alpha_1*\overline{\delta}][\delta*\alpha_2*\overline{\delta}][\delta*\alpha_1*\overline{\delta}]^2[\delta*\overline{\alpha_2}*\overline{\delta}]=A_1A_2A_1^2A_2^{-1}$ and $\varphi_{2*}[\gamma]=1$

Therefore, by Seifert Van Kampen, $\pi_1(X,x_0)=\langle A_1,A_2; A_1A_2A_1^2A_2^{-1}\rangle$

This is correct?


1 Answer 1


It is correct. In fact, let $Y = S^1_1 \vee S^1_2$ denote the wedge of two copies $S^1_i$ of $S^1$. This is the space you denote as "the figure $8$". You attach a two-cell to $Y$ as indicated in fig. 25.10 to obtain the space $X$. We have $\pi_1(Y) = \mathbb Z * \mathbb Z$ = free group with generators $A_1, A_2$, where $A_i$ is represented by the identity map $S^1 \to S^1_i \subset Y$.

The attaching map $\phi : S^1 \to Y$ represents the word $A_1A_2A_1^2A_2^{-1}$ which gives the presentation $\pi_1(X,x_0)=\langle A_1,A_2; A_1A_2A_1^2A_2^{-1}\rangle$.

For a more general approach to questions like that see my answer to Can this counter example disprove the statement?


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