Hi,
very simple question here. In Hatcher's 'Algebraic Topology' the diagram above is used to describe the projective plane as a $\Delta$-complex(see p.102). Later the 2-boundary maps are given by $\partial_2(U) = -a+b+c$ and $\partial_2(L)=a-b+c$. Can someone explain why e.g. $\partial_2(L)$ may not be $-a+b-c$? The way it is described by Hatcher we traverse the 2-simplex $U$ in counter clockwise order and $L$ in clock wise order. But is there a simple argument to see that we have to do so?
Best Answer
The reason is each simplex has a defined order given by the order inherited by the sub-simplices:
A simplex always looks the same. In case of the projective plane in the diagram, they are just arranged in a convenient way, such that the gluing map of the boundaries exactly is the one needed to result in the desired space.
In case of $\Delta^2$-simplices it is easy to see the orientation by just starting with the vertex that has two outgoing arrows. Call it $v_0$ for example. The vertex with two incoming arrows is the last one $v_2$.
So each simplex has a description eg. $[v_0 v_1 v_2 v_3 v_4]$
And you can simply apply the definition of the boundary map.
In case of a $\Delta^2$-simplex with vertices $[w_0 w_1 w_2]$ that would give:
$$\partial [w_0 w_1 w_2] = [w_1 w_2] - [w_0 w_2] + [w_1 w_2]$$