Fundamental group of $\mathbb R^2 \setminus \mathbb Z^2$ is non-trivial.

Question

Show that $\pi_1 \left (\mathbb R^2 \setminus \mathbb Z^2 \right )$ is non-trivial.

My Attempt $:$

We know that $\mathbb R^2 \setminus \{(0,0) \}$ deformation retracts to $S^1.$ Scaling down by any $r \gt 0$ we can deform the punctured plane to a circle of radius $r$ and the deformation is precisely given by the continuous map $H : \mathbb R^2 \setminus \{(0,0) \} \times I \longrightarrow \mathbb R^2 \setminus \{(0,0) \}$ defined by $(x,t) \mapsto (1 - t) x + t \frac {r x} {\|x\|},$ $x \in \mathbb R^2 \setminus \{(0,0) \}$ and $t \in I.$ One can easily show that $H$ is a strong deformation retract and consequently all the points on the circle $C_r$ centered at the origin and radius $r \gt 0$ remains fixed during the homotopy. Now fix some $0 \lt r \lt 1$ and consider the loop $\gamma$ in $\mathbb R^2 \setminus \{(0,0) \}$ which traverses the circle $C_r$ once in the anticlockwise direction. Then after deformation $\gamma$ maps to itself which is a non-trivial loop in $C_r$ as $\pi_1 (C_r) \cong \mathbb Z$ and $\gamma$ corresponds to $1 \in \mathbb Z.$ So $\gamma,$ thought of as a loop in $\mathbb R^2 \setminus \{(0,0) \},$ is non-trivial as well since $\pi_1 (C_r) \cong \pi_1 \left (\mathbb R^2 \setminus \{(0,0) \} \right ).$ Then it follows that $\gamma$ is non-trivial when considered it as a loop in $\mathbb R^2 \setminus \mathbb Z^2$ as $\mathbb R^2 \setminus \mathbb Z^2 \subseteq \mathbb R^2 \setminus \{(0,0) \}.$

Am I right with my reasoning or is there anything needs to be added? Any suggestion in this regard would be warmly appreciated.

Thanks for investing your valuable time on my question.

Answer 1

Your proof is absolutely correct. Here is a suggestion making it a bit more "elegant".

Let $X = \mathbb R^2 \setminus \mathbb Z^2$ and $C \subset X$ be the circle with center at the origin and radius $1/2$. Let $i : C \to X$ be the inclusion.

There is a retraction $r : X \to C$ given by $r(x) = \dfrac{x}{2\lVert x \rVert}$.

Then $r \circ i = id_C$ and therefore $$r_* \circ i_* = id$$ for the induced homomorphisms on the fundamental groups (as basepoint for both spaces $C,X$ take any $c_0 \in C)$. This makes it impossible that $\pi_1(\mathbb R^2 \setminus \mathbb Z^2) = 0$ because in that case we would get $r_* \circ i_* = 0$ which contradicts the fact that $\pi_1(C) \ne 0$.