Consider the following topological space $X$, which consists of a Klein bottle with two discs glued inside.
I am trying to compute the cellular homology of this space. Two approaches:
direct approach: let $f_1$ be the barrel of the Klein Bottle, $f_2$ and $f_3$ be the two discs, $e_1$ be the rim of the cylinder that is glued together to obtain the Klein Bottle, $e_2$ the seam of the cylinder, and $v_1$ the (only) vertex.
I obtain a chain complex $\langle f_1, f_2, f_3\rangle \xrightarrow{\begin{pmatrix}2 & 1 & 1\\ 0& 0 & 0\end{pmatrix}}\langle e_1, e_2 \rangle \xrightarrow{\begin{pmatrix}0 & 0\end{pmatrix}}\langle v_1\rangle$, which leads to betti numbers $(1, 1, 2)$ and $\chi(X)=2$ over $\mathbf F_2$.
using Mayer-Vietoris sequence: Cover $X$ by the Klein Bottle $K$ and by the sphere $S^2$ spanned by the two discs inside. We have $K\cap S^2\simeq S^1$, thus the MV-sequence is the exact sequence
$$\begin{matrix}&0&\to& \mathbf Z&\to&H_2(X)&\to\\ \to&\mathbf Z&\to&\mathbf Z\oplus\mathbf Z/2\mathbf Z&\to&H_1(X)&\to\\ \to&\mathbf Z&\to&\mathbf Z&\to&H_0(X)&\to & 0\end{matrix}$$
or, to take a field,
$$\begin{matrix}&0&\to& \mathbf F_2&\to&H_2(X)&\to\\ \to&\mathbf F_2&\to&\mathbf F_2^2&\to&H_1(X)&\to\\ \to&\mathbf F_2&\to&\mathbf F_2&\to&H_0(X)&\to & 0\end{matrix}$$
However, $\chi(K\cap S^2)=0, \chi(K)+\chi(S^2)=1$, thus we would have $\chi(X)=1$, which is not compatible with the direct approach.
Question: Where is my misconception?