A simple fundamental group I think I have this one, but I want to make sure:
Using Van Kampen's theorem,
Find the fundamental group of two disjoint spheres with each north pole identified, and each south pole identified.
The open sets I chose are sphere 1 minus the north pole, sphere 2 minus the south pole. The intersection is nonempty with trivial fundamental group since poles are identified, so we get trivial*trivial=trivial. Correct?
 A: Let $X$ be your space. Another possibility, not using van Kampen theorem (explicitely), is to glue a 3-cell inside each sphere. Because the fundamental group depends only on the 2-skeleton, $\pi_1(X)$ is isomorphic to the fundamental group of our new space $\tilde{X}$. Now, $\tilde{X}$ clearly retracts by deformation on a circle, hence $$\pi_1(X) \simeq \pi_1(\tilde{X}) \simeq \pi_1(\mathbb{S}^1) \simeq \mathbb{Z}.$$
A: Denote the two spheres by $S_1,S_2$. Let
$$A_i=S_i\cap\{(x,y,z)\mid y>-0.1\},\qquad B_i=S_i\cap\{(x,y,z)\mid y<0.1\}\\
A=(A_1\sqcup A_2)/\sim,\qquad B=(B_1\sqcup B_2)/\sim$$ 
$A, B$ are subspaces of $X$, and their intersection is homotopy equivalent to a
$$\Huge\Bbb O$$
Let us denote the four edges pointing downwards by $a,b,c,d$ left to right. Then 
$\pi_1(\Bbb O,N)$ is generated by $b^{-1}a,\ c^{-1}b,\ d^{-1}c$.
Let $γ_A,γ_B$ denote the class of $c^{-1}b$ in $\pi_1(A,N),π_1(B,N)$ respectively.
$A$ and $B$ can be thought of each as two disks with their north poles glued together and their south poles glued together. The fundamental group 
$π_1(A,N)$ is generated by $γ_A$, and $π_1(B,N)$ is generated by $γ_B$
Applying van Kampen we see that $π_1(X,N)$ is the free product $\pi_1(A,N)*\pi_1(B,N)$ modulo the smallest normal subgroup containing  $γ_A^{-1}γ_B$. 
$$π_1(X,N) = \langle γ_A,γ_B \mid γ_A^{-1}γ_B \rangle  \cong  \Bbb Z$$
In order to see that $⟨a,b ∣ a^{−1}b⟩\cong \Bbb Z$, we consider the homomorphism
$$\phi: ⟨a,b⟩ \to \Bbb Z,\quad a,b\mapsto 1$$
So $ϕ$ sends a word to the sum of the exponents appearing in it. Since $ϕ(a^{-1}b)=0$, the normal subgroup $N$ is in ker$(ϕ)$ and we have an induced morphism 
$\barϕ:⟨a,b ∣ a^{−1}b⟩\to \Bbb Z$. This $\barϕ$ is an isomorphisms if the kernel of $ϕ$ is precisely $N$. In practice this means we have to show that if some word represents $0$ in $\Bbb Z$, then it can be turned into the empty word by applying the relation $a=b$.
A: You may also try by collapsing a segment from the north pole to the south pole in one of the sphere, thus obtaining the wedge sum of a sphere (call it $S$) with a sphere with the two poles indentified (call it $T$), and this space you have obtained is homotopy equivalent to the former (essentially this is due to the fact that a segment is contractible). Now, since the basepoint (the point in which the glueing happens) has contractible neighbourhoods in both space, the fundamental group of your space is the coproduct of the fundamental group of $S$ and $T$, which is easily computable as being $\mathbb{Z}$.
A: First of all, we may picture the two spheres as a torus pinched in two points, namely North and South points of the two spheres.
Doing this, it's clear that our space $X$ is path connected and can be covered by two open sets: $A = \left[\left(S^2 \amalg S^2 \right)/ \sim \right] \setminus \left[\,\{N\}\,\right]$ where $\sim$ is the relation with North and South poles identified and $\left[\,\{N\}\,\right]$ is the class of equivalence of North poles of the two spheres; and $B$, the first half of the 2-pinched torus, namely $\left[ S^2 \amalg \{S\} \right]$.
$A$ and $B$ are obviously open sets and path connected; $A$ is homeomorphic to $S^2 \vee S^2$, $B$ is homeomorphic to $S^2$. Their intersection $A\cap B$ is homeomorphic to an infinite cylinder and thus its fundamental group is $\mathbb{Z}$.
Now, Van Kampen says that $\pi(X) = \pi(A) \times \pi(B)$ if $A \cap B$ is simply connected. In a more general version, it says that $\pi(X) = \left[ \pi(A) \times \pi(B) \right] \star \pi(A\cap B)$, where $\star$ is the free product. Now we know that $\pi(A)$ is trivial (again for Van Kampen, if you desire) and $\pi(B)$ is trivial. So, $\pi(X) = \pi(A\cap B) = \mathbb{Z}$.
