Examples of $\pi_1 (X) = \mathbb{Z}$ I want to know some examples of topological spaces whose fundamental group is isomorphic to set of integers. 
First, of course i know $\mathbb{S}^1$, and its deformation retract, $\mathbb{R}^2 - \{ 0, 0\}$, $\mathbb{C}^1 - \{0 \}$, etc, are isomorphic to set of integers. 
And of course from Van kampen theorem, attaching simply connected spaces, or by wedge sum of simpily connected space for  $\mathbb{S}^1$, etc, we can produces $\mathbb{Z}$.   
Can you give me some other examples than that?
 A: Take the loop space $\Omega(S^2)$.
A: One interesting example is the pseudocircle, which is the finite topological space $V=\{P_1,P_2,L_1,L_2\}$ where the open sets are those subsets $U$ such that if $P_1\in U$ or $P_2\in U$ then $L_1$ and $L_2$ are in $U$.  My notation is suggestive; you can think of this space as a circle as follows

Then the open sets are precisely the sets that 'look' open in this diagram; i.e.:
$$\emptyset, \{L_1\}, \{L_2\}, \{L_1, L_2\}, \{P_1, L_1, L_2\}, \{P_2, L_1, L_2\}, \{P_1, P_2, L_1, L_2\}$$
It turns out that this space has fundamental group $\mathbb Z$.  
In fact, we can say something stronger.  The pseudocircle is weakly homotopy equivalent to the circle $S^1$.  What that means is that there is a continuous function $S^1\to V$ such that the induced morphisms on the fundamental group and on all higher homotopy groups are isomorphisms.  So the pseudocircle shares with $S^1$ the property that all higher homotopy groups are $0$.  
A space need not have zero higher homotopy groups in order to have fundamental group $0$, though.  For example, note that the homotopy group functors all preserve products and that $\pi_2(S^2)=\mathbb Z$.  That means that
\begin{align}
\pi_1(S^1\times S^2)&=\pi_1(S^1)\times\pi_1(S^2)=\mathbb Z\times 0=\mathbb Z\\
\pi_2(S^1\times S^2)&=\pi_2(S^1)\times\pi_2(S^2)=0\times \mathbb Z=\mathbb Z
\end{align}
So $S^1\times S^2$ is an example of a topological space which is definitely different from $S^1$ from the point of view of algebraic topology, but which still has fundamental group $\mathbb Z$.  
A: The quotient space $X = (\Bbb R \times \{0,1\}) / {\sim}$ where we identify $(x,0)\sim(x,1)$ for each $x\ne0$. This is called the real line with two origins, and it is not Hausdorff.
We can cover $X$ with two open subsets $U$ and $V$, corresponding to $\Bbb R\times\{0\}$ and to $\Bbb R \times \{1\}$, respectively. Since their intersection $U\cap V$ has two components, one for positive numbers and one for negative numbers, we have to choose two base points, say $a=[(1,0)]$ and $b=[(-1,0)]$. Both subspaces have the fundamental groupoid $\mathbf I$ which consists of one arrow $i:a\to b$ and its inverse. The van Kampen theorem for groupoids yields as the fundamental groupoid $\pi_1(X,\{a,b\})$ that groupoid $G$ which is obtained by gluing two copies of $\mathbf I$ at their points $a$ and at their points $b$.  The group at one point of $G$ is isomorphic to $\Bbb Z$ and this is the fundamental group of $X$.
A: Take any (path connected) base-pointed space $X$ whose fundamental group has a subgroup isomorphic to $\mathbb{Z}$. Take a base-pointed topological space $Y$ and a covering map $Y \to X$ that corresponds to that subgroup, under the bijective correspondence between subgroups and pointed covering spaces. It follows that $\pi_1 Y \approx \mathbb{Z}$.
A: Two examples of particular importance in symplectic topology are the Lie groups of unitary matrices $U(n)$ and symplectic matrices $Sp(2n, \mathbb{R})$. The symplectic matrices are the real $2n \times 2n$ matrices that satisfy $A^T J A = J$, where 
$$
J = \begin{pmatrix}
0 & -I_n \\
I_n & 0 
\end{pmatrix}
$$
is the standard complex structure on $\mathbb{R}^{2n}$.
In the case of $U(n)$, an isomorphism of $\pi_1 (U(n))$ with $\mathbb{Z}$ is induced by $\det : U(n) \to S^1$.
Another Lie group example is $GL(n,\mathbb{C})$. One can show via polar decomposition that it actually deformation retracts onto $U(n)$. 
A: The Möbius strip, or more generally anything that deformation retracts to $S^1$. 
For a less trivial example, take an infinite wedge of circles indexed by integers and for each triple of integers $x,y,z$ such that $x+y=z$, glue a 2-cell with boundary $x+y-z$. The fundamental group is generated by "1". (This correspond to the 2-skeleton of a common model of the Eilenberg-Maclane space $K(\Bbb Z,1)$ and this 2-skeleton has a nontrivial second homotopy group).
