Euler characteristic of $n$-dimensional disk $D^n$ seen as a CW-complex is -1?

Question

I know that the Euler characteristic of the closed $n$-dimensional disk $D^n$ is $\chi(D^n)=1$ (see e.g. this question).

On the other hand, $D^n$ can be seen as a CW-complex consisting of only a single $n$-cell. According to the definition of the Euler characteristic as the alternating sum of its number of $k$-cells (see e.g. here on Wikipedia), this means that $$ \chi(D^n) = 0 - 0 + \dots + (-1)^n = (-1)^n, $$

which of course is incompatible with above result. So where is the mistake?

Answer 1

In a CW complex, each $k$-cell has to be attached to the $(k-1)$-skeleton, building up from the $0$-skeleton, which is a discrete collection of points. A $k$-cell is topologically a disk $D^k$ and its boundary is the sphere $S^{k-1}$. The attaching map has to take these boundary points to the $(k-1)$-skeleton, consisting of cells of dimension at most $k-1$. What is the top dimensional cell attached to in your proposed complex?