# Problem 25.1 (b) Kosniowski. Computing the fundamental group of a pentagonal region.

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.

Solution:

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?

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?