Mistake in computing boundary maps in $S^n$ I am giving $S^n$ the CW complex structure with $2$ $k-$cells in each dimension $0\leq k\leq n$ . The attaching maps are given by the pushout diagram. 
$\require{AMScd}
\begin{CD}
S^{k-1}\cup S^{k-1} @>{i}>> D^k\cup D^k\\
@V{id\cup id}VV @VVV \\
S^{k-1} @>>> S^k
\end{CD}
$
I want to compute the chain complex $C^{CW}_*(S^n)$
Since there are $2$ cells in each dimension we have $C^{CW}_k(S^n)=\mathbb Z^2 \ ; 0\leq k\leq n$
Now for the boundary maps.
Suppose $$d:C^{CW}_k(S^n)\rightarrow C^{CW}_{k-1}(S^n)$$ is given by
$$D^k_i\mapsto n_{1i}D_1^{k-1} + n_{2i}D_2^{k-1} \ ; i=1,2$$
Then $$n_{ij} = deg \ (S_i^{k-1} \rightarrow S^{k-1} \rightarrow S^{k-1}/S^{k-2}\rightarrow S^k_j) $$
The map in the brackets is either quotienting out the upper hemisphere or the lower hemisphere and hence each has degree $1$ (See my previous question Calculating degree of a map )
Thus we get $$ d(1,0)=(1,1)$$ $$d(0,1)=(1,1)$$ 
But $d^2=0$ and in this case it is clearly non zero. I definitely have made a mistake somewhere in computing the degree. Can someone please point it out?
 A: One thing that is glossed over when the boundary map is being described is that just because we have a map $S^n \rightarrow X $ where $X$ is homeomorphic to $S^n$, does not mean there is a sensible notion of degree for this map. First you must pick a specific homeomorphism, and then you can work with degress.
What is important is that we choose this homeomorphism in a consistent manner. So in the case of cellular homology we want to talk about the degree of a map $S^n \rightarrow X^{n} \rightarrow X^n / (X^n - e^n_\alpha)$. What we can do to consistently talk about a degree map, even as $\alpha$ varies, is to identify $D^n$ with its image in $X^n$, then pick some homeomorphism from $D^n / \partial D^n \rightarrow S^n$.
Now let's say we are trying to compute some boundary maps for $S^2$ with your decomposition. We can see that some unexpected issues might come up:
For example, if we call the $0$-cells $A$ and $B$, which endpoints of $[0,1]$ did we attach where? Say that for one $1$-cell, $U$, we attached them one way and for the other, $V$, we attached them the other way. If you imagine including $S^1$ by quotienting out $U$ versus including $S^1$ by quotienting out $V$, you will see one has degree $1$ and one has degree $-1$ (you should be able to pick a homeomorphism so that one looks like the identity and one looks like a reflection).
However, you might think that if you attach them both in the same way the degrees must be the same. However, this is not the case because when you are considering the attaching map $S^1 \rightarrow X^1$, you will be attaching left to right on one copy of $[0,1]$ and right to left on the other copy. This will cause the degrees to again be opposite.
These opposing signs will be what cause the square of the boundary to be $0$.
