1. ## Hausdorff space, structure

Let $\displaystyle X$ be a Hausdorff space such that $\displaystyle X= A \cup B$ where $\displaystyle A$ and $\displaystyle B$ are each homeomorphic to a torus, and $\displaystyle A \cap B =\{x_0\}$.

What is the structure of $\displaystyle \pi_1(X, x_0)$?

How do I describe the structure of this topological space's fundamental group? I know how to draw it, etc. But, I don't know much about this space's fundamental group.

2. Originally Posted by Erdos32212
Let $\displaystyle X$ be a Hausdorff space such that $\displaystyle X= A \cup B$ where $\displaystyle A$ and $\displaystyle B$ are each homeomorphic to a torus, and $\displaystyle A \cap B =\{x_0\}$.

What is the structure of $\displaystyle \pi_1(X, x_0)$?

How do I describe the structure of this topological space's fundamental group? I know how to draw it, etc. But, I don't know much about this space's fundamental group.
Lemma 1. Let $\displaystyle X = U \cup V$, where U and V are open sets of X. If $\displaystyle U \cap V$ is simply connected, then $\displaystyle \pi(X)$ is the free product of groups $\displaystyle \pi(U)$ and $\displaystyle \pi(V)$ with respect to the homomorphisms $\displaystyle \phi_1 : \pi(U) \rightarrow \pi(X)$ and $\displaystyle \phi_2 : \pi(V) \rightarrow \pi(X)$.

If A and B were open subsets of X, we could apply Lemma 1 with U=A and V=B to determine the structure of $\displaystyle \pi_1(X, x_0)$.

Alternatively, we need to choose two open sets U and V such that A and B are deformation retracts of U and V, respectively.

Choose the circles and lines $\displaystyle S_1, l_1 \in A$ and $\displaystyle S_2, l_2 \in B$ such that if you remove the $\displaystyle S_1, l_1$ from A and $\displaystyle S_2,l_2$ from B, A and B should become simply connected (Think of a reverse procedure of making a torus from a rectangle paper).

Let $\displaystyle U=X - S_2 -l_2$ and $\displaystyle V=X - S_1 -l_1$, where $\displaystyle x_0$ does not belong to any $\displaystyle S_1, S_2, l_1$ and $\displaystyle l_2$. Then, A and B are deformation retracts of U and V, respectively

Now, we have $\displaystyle X = U \cup V$, and $\displaystyle U \cap V = X -S_1 -S_2 -l_1 -l_2$, where $\displaystyle U \cap V$ is contractible.

Then, by lemma 1, we conclude that $\displaystyle \pi_1(X, x_0)$ is the free product of groups $\displaystyle \pi_1(U, x_0)$ and $\displaystyle \pi_1(V, x_0)$, or the free product of groups $\displaystyle \pi_1(A, x_0)$ and $\displaystyle \pi_1(B, x_0)$.

Since $\displaystyle \pi_1(A) \cong Z \times Z$ and $\displaystyle \pi_1(B) \cong Z \times Z$, $\displaystyle \pi_1(X, x_0)$ is isomorphic to the free product of groups $\displaystyle Z \times Z$ and $\displaystyle Z \times Z$.