Van Kampen's Theorem with Torus and Projective Plane I'm having some trouble finding the sets $U$ and $V$ to use for this problem.
Let $T = S^1 \times S^1$ be the torus and $P=S^2/(x \sim -x)$ the projective plane. Form the space $X$ by identifying the circle $S^1 \times \{ x_0 \}$ in $T$ with the image of the equator in $P$ by a homeomorphism. Find a presentation of $\pi_1(X)$ using the van Kampen theorem.

I will use Theorem 10.1 and 10.3 in Lee's book, with $U$ and $V$ as in the comments. Then recall that 
\begin{align*}
\pi_1(U) &= \langle x,y\;|\; xyx^{-1}y^{-1}\rangle \simeq \mathbb{Z}\times \mathbb{Z}\\
\pi_1(V) &= \langle z\;|\;z^2 = 1\rangle \simeq \mathbb{Z}_2 \\
\pi_1(U\cap V) &= \langle w \;|\; \varnothing\rangle \simeq \mathbb{Z}
\end{align*}
By Theorem 10.1, calling our main space $X$, we have
$$
\pi_1(X) \simeq \pi_1(U) \ast_{\pi_1(U\cap V)} \pi_1(V) 
$$
and then Theorem 10.3 gives us a way to actually write out this amalgamated product:
$$
\pi_1(X) \simeq \pi_1(U) \ast_{\pi_1(U\cap V)} \pi_1(V)  \simeq \langle x,y,z\;|; xyx^{-1}y^{-1}, z^2, z=x^2\rangle
$$
EDIT: Theorem 10.1 is just the statement of SVK and Theorem 10.3 is the following:
[Presentation of an Amalgamated Free Product] Let $f_1:H \to G_1$ and $f_2:H \to G_2$ be group homomorphisms. Suppose $G_1, G_2$ and $H$ have the following finite presentations:
\begin{align*}
G_1 &\simeq \langle \alpha_1,\ldots, \alpha_m \;|\;\rho_1,\ldots, \rho_r\rangle ;\\
G_2 &\simeq \langle \beta_1,\ldots, \beta_n\;|\; \sigma_1,\ldots, \sigma_s\rangle ;\\
H&\simeq \langle \gamma_1,\ldots, \gamma_p\;|\; \tau_1,\ldots, \tau_t\rangle 
\end{align*}
Then the amalgamated free product has the presentation
$$
G_1 \ast_H G_2 \simeq \langle \alpha_1,\ldots, \alpha_m,\beta_1,\ldots, \beta_b\;|\; \rho_1,\ldots, \rho_r,\sigma_1,\ldots,\sigma_s, u_1=v_1,\ldots,u_p=v_p\rangle,
$$
where $u_a$ is an expression for $f_1(\gamma_a) \in G_1$ in terms of the generators $\lbrace\alpha_i\rbrace$ and $v_a$ similarly expresses $f_2(\gamma_a) \in G_2$ in terms of $\lbrace \beta_j\rbrace$. 
