= Solution
Cut the square model of the <Klein bottle> along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a <Möbius band>, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The <Seifert-van Kampen theorem> therefore gives
$$
\boxed{
\pi_1(K,k_0)
\cong\langle a,b\mid a^2b^{-2}=1\rangle
}.
$$
Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentation
$$
\langle x,y\mid yxy^{-1}=x^{-1}\rangle.
$$
The orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroup
$$
H=\langle a^2,a^{-1}b\rangle\cong\mathbb Z^2.
$$
By the classification of connected covering spaces, $H$ determines the <orientation double cover of the Klein bottle>
$$
p:(T^2,x_0)\longrightarrow(K,k_0).
$$
Choosing torus generators $\alpha,\beta$ along the two translation directions, we may take
$$
\boxed{
p_*(\alpha)=a^2,
\qquad
p_*(\beta)=a^{-1}b
}.
$$
These commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.
The space $Y$ is the <mapping cylinder> of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/{\sim}\cong K$. Hence
$$
\boxed{
\pi_1(Y,y_0)\cong\pi_1(K,k_0)
\cong\langle a,b\mid a^2b^{-2}\rangle
}.
$$
Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, to
$$
X=U\cup(X\setminus K).
$$
The intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to $\pi_1(U)\cong\pi_1(K)$ has image precisely
$$
p_*\pi_1(T^2)=H.
$$
The orientation homomorphism
$$
\pi_1(U)\longrightarrow\pi_1(U)/H\cong C_2
$$
is zero on this intersection, so it is compatible with the trivial homomorphism from $\pi_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjection
$$
\pi_1(X)\twoheadrightarrow C_2.
$$
Therefore
$$
\boxed{X\text{ cannot be simply connected}}.
$$
This is the <Klein-bottle mapping-cylinder obstruction to simple connectivity>.
Back to article page
