$X = D^{2} \times S^{1} \cup_{f} S^{1} \times D^{2}$, where $f : S^{1} \times S^{1} \to S^{1} \times S^{1}$ is a map induced by the linear map on $\mathbf{R}^{2}$ given by the matrix $$\left( \begin{array} {cc} a & b \\ c & d \end{array} \right)$$ where $a, b, c, d$ are integers. How can I represent the fundamental group $\pi_{1}(X)$ in terms of $a, b, c, d$? Please give me some hints.
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Hint: Google "lens spaces". – Moishe Kohan May 13 '14 at 00:54
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@studiosus Thanks, I got it $\pi_{1}(X)$ is $\mathbb{Z}_{p}$ for some $p$ when the matrix is invertible. But what should I do if the rank of matrix is 1 or 0? – fiverules May 13 '14 at 01:47
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Hint: $(D^2\times S^1)\cap(S^1\times D^2)$ is a torus which we know the fundamental group of. We also know the fundamental groups of each component of the adjunction. Try using Van-Kampen's theorem - the amalgamation of the free product will be dependent on the map $f$ so be careful when considering the quotient.
Chris Brooks
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Dan Rust
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1I first think that when if a=1, b=0, c=0, d=1, then by Van Kampen, we get $((\mathbb{Z}) \times (\mathbb{Z}))/(\mathbb{Z} \times \mathbb{Z}) = 0$, but I don't know how to describe the free product explicitly when $a, b, c, d $ varies. – fiverules May 13 '14 at 00:21