Computing the homology of a subspace of $R^3$ Let $P_1 ,...,P_5$ be mutually distinct planes in $\Bbb{R}^3$ such that:


*

*the intersection of any distinct 2 is a line.

*the intersection of any distinct 3 is a point.

*the intersection of any distinct 4 is an empty set.


Let $X=\bigcup_{i=1} ^5 P_i$ with the topology as a subspace of $\Bbb{R}^3$
Given that $\chi (X)=5$ and that $X$ is simply connected, how do you compute its homology with integral coefficients?

The space is a CW-complex of dimension $2$, so there are only $H_0$, $H_1$ and $H_2$ to care for. Since it is connected, $H_0\cong\mathbb Z$ and since you say it is simply connected, $H_1=0$. It follows that $H_2$ must be an abelian group of rank $4$ if the Euler characteristic is to be $5$, as you claim it is.
So you are left with finding the possible torsion in $H_2$.

Consider the plane $P_1$ and the four lines $P_1\cap P_i$ with $2\leq i\leq 5$. They intersect in pairs and no three are concurrent. They divide the plane in 10 components, 7 of which are unbounded. Of course, the same picture shows up in each of the five planes.
Let $Z$ be the subspace of $X$ which is the union of all the bounded pieces in the five planes, and let $Y$ be the subspace of $X$ which is the union of $Z$ and the ten lines which appear as pairwise intersections of the $5$ planes. Then $X$ deformation-retracts onto $Y$ (one can deform each of the unbounded $2$-dimensional cells onto their boundaries) and $Y$ deformation-retracts onto $Z$ (one can contract each outgoing half-ray to its vertex. (Of course, one should draw pictures of all this to make sense of it :-) )
Now view $\mathbb R^3$ as a subset of $S^3$ (one-point compactification). Then $Z$ is a subspace of $S^3$. Alexander duality tells us that $\tilde H_2(Z)\cong\tilde H^0(S^3\setminus Z)$ (here the tildes mean reduced (co)homology) Since $\tilde H^0(S^3\setminus Z)$ is a free group, this implies $H_2(X)=H_2(Z)=\tilde H_2(Z)$ is free.
The previous observations, then, imply that $H_2(X)\cong\mathbb Z^4$.

Notice that the $4$ is one less than the number of components of $S^3\setminus Z$. One can compute this directly, without the hypothesis that $\chi(X)=5$. Indeed, there is a well-known formula for the number of bounded components in the complement of a arrangement of hyperplanes due to Zavlasky, which gives this number in terms of the intersection poset of the arrangement, and here it is completely given. This is explained, for example, in this very nice notes by Richard Stanley on arrangements of hyperplanes.
Also, Alexander duality also gives $H_1(X)\cong H^1(S^3\setminus Z)$. Now $S^3\setminus Z$ is the disjoint union of the «big component», that is, the unbounded component of $Z$, and the bounded components in the arrangement of hyperplanes: all these are contractible, so $H^1(S^3\setminus Z)$ is zero. This means we can also dispense with the hypothesis that $X$ is simply connected.

All this will work for a general arrangement of hyperplanes in $\mathbb R^n$, of course. The conclusion will be that the union of the hyperplanes has the homology of a wedge of spheres, as many as there are bounded components in the complement of the arrangement. In fact,I suspect the space is actually homotopically equivalent to such a wedge of spheres. This should be well-known by the people who know such things —I'd look in the book by Orlik and Terao.
