Let $ U = \mathbb{S}^{1} \times \mathbb{S}^{1} - interior\ of\ a\ disc$ and $V=\mathbb{K} - interior\ of\ a\ disc$, where $\mathbb{K}$ is the Klein bottle. Let $f : \partial U \to \partial V$ be a function defined by $f(e^{i\theta})=e^{ni\theta}$, where $n$ is a positive integer. Compute $\pi_{1}(M)$ where $M = U \cup V$ where $\partial U$ is identified with $\partial V$ via $f$.
It's like the connected sum of torus and klein bottle but rather than gluing the boundaries of the discs, $\partial V$ is identified with $\partial U$ by twisting it n times around. Using Van Kampen Theorem seems to be the way to go but I don't know what to do with the identification map. Thanks in advance.
Yes, this can be done by the Seifert-van Kampen theorem. Thicken $U$ and $V$ inside $M$ to get an open cover $\{\mathcal{U}, \mathcal{V}\}$ of $M$ where each open set deformation retracts to $U$ and $V$ respectively. $\mathcal{U} \cap \mathcal{V}$ deformation retracts to $\partial U = \partial V = S^1$.
According to SvKT theorem, $\pi_1(M) \cong \pi_1(\mathcal{U}) * \pi_1(\mathcal{V})/\langle i_{\mathcal{U}} i_{\mathcal{V}}^{-1} \rangle$ where $i_{\mathcal{U}}$ is the map on $\pi_1$ induced by the inclusion $\pi_1(\mathcal{U} \cap \mathcal{V}) \to \pi_1(\mathcal{U})$ and similar for $i_\mathcal{V}$.
Note that $U$ deformation retracts to the figure eight, and $\partial U$ represents the loop $aba^{-1}b^{-1}$ in the fundamental group of the figure eight under the deformation retract, where $a$ and $b$ are generator loops in the $\pi_1$ of figure eight. Similarly, $V$ deformation retracts to the figure eight, and $\partial V$ represents the loop $cdcd^{-1}$ in the fundamental group of the figure eight.
Now, we are gluing $U$ and $V$ by the degree $n$ map $\partial U \to \partial V$. Thus, $i_\mathcal{U}$ sends the generator $1$ of $\pi_1( \mathcal{U} \cap \mathcal{V}) \cong \Bbb Z$ to $aba^{-1}b^{-1}$ in $\pi_1(\mathcal{U}) \cong \langle a,b\rangle$ and $i_\mathcal{V}$ sends the generator $1$ of $\pi_1(\mathcal{U} \cap \mathcal{V}) \cong \Bbb Z$ to the element $(cdcd^{-1})^n$ in $\pi_1(\mathcal{V})\cong \langle c, d\rangle$.
Thus, $\pi_1(M) \cong F_4/\langle aba^{-1}b^{-1}(cdcd^{-1})^{-n} \rangle \cong \langle a, b, c, d | aba^{-1}b^{-1} = (cdcd^{-1})^n \rangle$.