How would I construct the CW-complex of a triangle?
My thoughts:
$X^0$ is 3 copies of $e_0$.
$X^1$: we take three copies of $e_1$, and each one gets its boundary identified with two of the $e_0$s. So now we have a triangle with an empty inside.
$X^2$: we take a single $e_2$ and identify its boundary ($S^1$) with the boundary of the triangle, thus 'filling' it in.
However, my lecture notes show this:
Why is there only one point in $X^0$? I understand how they've got $X^1$ from $X^0$, but not how they've got $X^2$ from $X^1$. What is that map $\phi$ of degree $3$, and how it give a triangle?
The CW-complex being constructed in your notes is not a triangle (and your approach to constructing a triangle as a CW-complex is perfectly correct). Rather, it is a triangle with the three edges of the triangle all glued together, with the orientation indicated by the arrows in the picture. There is only one vertex, because when you glue the edges like that, all three vertices end up glued together. There is only one edge, because you've glued all three together, and it goes from the single vertex to itself. Finally, the attaching map for the two cell is the "boundary of the triangle", except that the three edges of the triangle are all identified, so going around the boundary of the triangle is the same as going along the one edge three times in the same direction. This is a map $S^1\to S^1$ of degree $3$.