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Compute the fundamental group of the following region at any base point. enter image description here

Note that we've excluded the open disk enclosed by the blue circle.

The region is certainly not convex. I tried finding an obvious deformation retract, but couldn't. Any help would be appreciated.

Edit: So first I was under the impression that the region was simply connected (which as pointed out by Martin R. seems intuitively incorrect), hence the change in question. I was wondering if it would be possible to compute the fundamental group of this region ... in particular I am starting to suspect that it should have the blue circle as a deformation retract (whereupon its fundamental group should be isomorphic to $\mathbb Z$), but I can't construct any deformation retraction (or prove rigorously that it is/is not simply connected).

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    $\begingroup$ You cannot prove it because it is not true. $\endgroup$ Commented May 21, 2020 at 3:51
  • $\begingroup$ Really? What is your reasoning (In particular, do you think the circle is a deformation retract)? $\endgroup$ Commented May 21, 2020 at 4:00
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    $\begingroup$ Because it has a hole. A circular path around that hole cannot be contracted to a point. Or am I misunderstanding something? $\endgroup$ Commented May 21, 2020 at 4:01
  • $\begingroup$ I suppose you are right, that is the reason why I felt that it might deformation retract onto the blue circle as well. $\endgroup$ Commented May 21, 2020 at 4:08

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This region has the property that it is radially convex (from the center of the omitted circle). So one can construct a pretty simple deformation retract. For simplicity I'll take the center of the omitted circle to be the origin and its radius to be $1$. If $R$ is the region in the picture, define the function $f\colon R\times[0,1]\to R$ by the following formula in polar coordinates: $$ f(re^{i\theta},t) = r^te^{i\theta}. $$ Then $f(z,1)$ is the identity function, $f(z,0)$ has image lying in the unit circle, and $f$ is continuous in both variables.

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    $\begingroup$ Another option, avoid polar coordinates, is just linearly scaling the norm: $f(p,t) = \left(1 + t(\frac{1}{|p|} - 1)\right)p$. $\endgroup$ Commented May 21, 2020 at 21:39

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