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Lets say $X$ is a path connected space. Now we know that any loop on any point $x_0$ can be transferred over to any other point $x_1$. It is claimed that $\pi_1(X)=0$. I understood that there is only one homotopy class here, but what I don't understand is why $\pi_1(X)=0$. We can design the space $X$ in such a way that the origin is not part of it. Still the book (Algebraic Topology by Hatcher) claims it is zero. Why? Also why it is just a single number. Also take the following instance.

I can design a path connected space which is NOT nullhomotopic, e.g. a square with a hole in the middle. Then not every loop can be shrunk to a point, hence just a single point can not be part of homotopy class.

What I am getting wrong here.

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The fundamental group of a path connected space is not necessarily trivial. What Hatcher says is given a path connected space then the fundamental group is independent of the base point you consider.

In your example you can homotopy the square with a puncture to $S^1$ which has fundamental group $\mathbb{Z}$ (homotopy equivalence preserves fundamental group).

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  • $\begingroup$ "you can homotopy the square with a puncture to $S^1$ which has fundamental group $\mathbb{Z}$" - how? I was also thinking, but what could be the construction? $\endgroup$ – user3001408 May 29 '17 at 13:25
  • $\begingroup$ Just think about it this way: you can compress and bend the boundary of the quake to that of a disk, and then you stretch out the puncture of the disk to the boundary. Then you are left with a circle. (All of these moves are allowed for homotopy equivalence). $\endgroup$ – ADA May 29 '17 at 13:30

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