Suppose that $n\ge 2$. We previously noted that ${\mathbb{R}}^{n}\setminus \{0\}$ is homotopy equivalent to ${S}^{n-1}$ and so has ${H}_{0}\simeq {H}_{n-1}\simeq \mathbb{Z}$, with all other homology groups zero. One might ask what happens if we remove several points ${a}_{1},\mathrm{\dots},{a}_{m}$ from ${\mathbb{R}}^{n}$ instead of just removing the origin. One can answer this question by induction on $m$, but to make the induction work smoothly, it is convenient to generalise slightly. Instead of just considering ${\mathbb{R}}^{n}\setminus \{{a}_{1},\mathrm{\dots},{a}_{m}\}$, we will consider $W\setminus \{{a}_{1},\mathrm{\dots},{a}_{m}\}$, where $W$ is an arbitrary convex open subset of ${\mathbb{R}}^{n}$. We first need the following lemma:

Suppose that $X\mathrm{=}U\mathrm{\cup}V$, where $U$ and $V$ are open and connected, and $U\mathrm{\cap}V$ is contractible. Then ${H}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathbb{Z}$ and ${H}_{k}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}{H}_{k}\mathit{}\mathrm{(}U\mathrm{)}\mathrm{\oplus}{H}_{k}\mathit{}\mathrm{(}V\mathrm{)}$ for all $k\mathrm{\ge}\mathrm{1}$.

This follows from Proposition 15.2 (the truncated Mayer-Vietoris sequence). That result includes the fact that ${H}_{0}(X)=\mathbb{Z}$, and that there are exact sequences

$${H}_{k}(U\cap V)\to {H}_{k}(U)\oplus {H}_{k}(V)\to {H}_{k}(X)\to {H}_{k-1}(U\cap V)$$ |

for all $k\ge 2$, and also an exact sequence

$${H}_{1}(U\cap V)\to {H}_{1}(U)\oplus {H}_{1}(V)\to {H}_{1}(X)\to 0.$$ |

As ${H}_{j}(U\cap V)=0$ for all $j>0$, these exact sequences show that the map ${H}_{k}(U)\oplus {H}_{k}(V)\to {H}_{k}(X)$ is an isomorphism for all $k\ge 1$. ∎

Let $W$ be a convex open subset of ${\mathbb{R}}^{n}$ (with $n\mathrm{\ge}\mathrm{2}$). Let $A$ be a finite subset of $W$, with $\mathrm{|}A\mathrm{|}\mathrm{=}m$ say, and put $X\mathrm{=}W\mathrm{\setminus}A$. Then ${H}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathbb{Z}$ and ${H}_{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}{\mathbb{Z}}^{m}$ and ${H}_{k}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathrm{0}$ for $k\mathrm{\notin}\mathrm{\{}\mathrm{0}\mathrm{,}n\mathrm{-}\mathrm{1}\mathrm{\}}$.

If $m=0$ then $X=W$ so $X$ is contractible so the claim is clear. Suppose instead that $m=1$ so $A=\{a\}$ say. Define $r:X\to {S}^{n-1}$ by $r(x)=(x-a)/\parallel x-a\parallel $. Next, as $W$ is open we can choose $\u03f5>0$ such that the open ball of radius $\u03f5$ around $a$ is contained in $W$. We can then define $j:{S}^{n-1}\to X$ by $j(y)=a+\u03f5y/2$. We find that $rj$ is the identity, and that $jr$ is homotopic to the identity by a straight-line homotopy, so $X$ is homotopy equivalent to ${S}^{n-1}$; the claim follows from this.

We now suppose that $m>1$ and that we have already proved the claim for sets of size less than $m$. We have only a finite number of vectors $a-{a}^{\prime}$ with $a,{a}^{\prime}\in A$ and $a\ne {a}^{\prime}$; choose any unit vector $u$ that is not perpendicular to any of these. We then find that the dot products $u.a$ (for $a\in A$) are all different, so we can list the elements of $A$ as ${a}_{1},\mathrm{\dots},{a}_{m}$ with $$. Choose constants $p,q$ with $$. Put $$ and $U=\{x\in X|u.x>p\}$, so $U$ and $V$ are open with $U\cup V=X$. The space $V$ is obtained by removing $\{{a}_{1},\mathrm{\dots},{a}_{m-1}\}$ from the convex set $$, so the homology of $V$ is given by the induction hypothesis. The space $U$ is obtained by removing ${a}_{m}$ from the set $\{w\in W|u.w>p\}$, so the homology of $U$ is given by the case $m=1$. The intersection $U\cap V$ is just the convex set $$ (with no points removed), so it is contractible. Lemma 20.1 therefore gives ${H}_{0}(X)=\mathbb{Z}$ and ${H}_{k}(X)={H}_{k}(U)\oplus {H}_{k}(V)$ for all $k>0$, and the induction step is clear from this. ∎

Let ${L}_{n}$ be the union of $n$ adjacent squares arranged horizontally. The case $n=3$ is illustrated below.

Let $U$ be the space obtained by removing the rightmost vertical edge; this is easily seen to be homotopy equivalent to ${L}_{n-1}$. Let $V$ be the space obtained by removing ${L}_{n-2}$, leaving just the rightmost square with two extra edges attached; this is easily seen to be homotopy equivalent to ${S}^{1}$. The intersection $U\cap V$ is a sideways H shape, and is easily seen to be contractible. We can therefore apply Lemma 20.1 to see that ${H}_{k}({L}_{n})={H}_{k}({L}_{n-1})\oplus {H}_{k}({S}^{1})$ for all $k>0$. It follows inductively that ${H}_{0}({L}_{n})=\mathbb{Z}$ and ${H}_{1}({L}_{n})={\mathbb{Z}}^{n}$ and ${H}_{k}({L}_{n})=0$ for $k>1$. Alternatively, we can let ${a}_{p}$ be the centre of the $p$’th square, so ${L}_{n}\subseteq {\mathbb{R}}^{2}\setminus \{{a}_{1},\mathrm{\dots},{a}_{n}\}$. It can be shown that the inclusion $i:{L}_{n}\to {\mathbb{R}}^{2}\setminus \{{a}_{1},\mathrm{\dots},{a}_{n}\}$ is a homotopy equivalence. The homology of ${L}_{n}$ can therefore be obtained from Proposition 20.2; it is easy to see that this gives the same answer.

Let ${W}_{n}$ consist of $n$ circles joined together at a single point. We can write this as $U\cup V$ as illustrated below. Given this, essentially the same argument as in Example 20.3 gives ${H}_{0}({W}_{n})=\mathbb{Z}$ and ${H}_{1}({W}_{n})={\mathbb{Z}}^{n}$ and ${H}_{k}({W}_{n})=0$ for $k\ge 2$.

The homology of the torus $T\mathrm{=}{S}^{\mathrm{1}}\mathrm{\times}{S}^{\mathrm{1}}$ is given by ${H}_{\mathrm{0}}\mathit{}\mathrm{(}T\mathrm{)}\mathrm{=}{H}_{\mathrm{2}}\mathit{}\mathrm{(}T\mathrm{)}\mathrm{=}\mathbb{Z}$ and ${H}_{\mathrm{1}}\mathit{}\mathrm{(}T\mathrm{)}\mathrm{=}{\mathbb{Z}}^{\mathrm{2}}$ and ${H}_{k}\mathit{}\mathrm{(}T\mathrm{)}\mathrm{=}\mathrm{0}$ for $k\mathrm{\ge}\mathrm{3}$.

We can form the torus by taking a square, gluing the top to the bottom and the left edge to the right edge. This is illustrated by the left hand picture below. The other pictures show two open sets $U,V\subseteq T$ with $U\cup V=T$.

The set $V$ is contractible, so ${H}_{0}(V)=\mathbb{Z}$ and ${H}_{k}(V)=0$ for $k>0$. The set $U\cap V$ is homotopy equivalent to ${S}^{1}$, so ${H}_{0}(U\cap V)=\mathbb{Z}$ and ${H}_{1}(U\cap V)=\mathbb{Z}.[u]$ and ${H}_{k}(U\cap V)=0$ for $k>1$. It might appear that $U$ is also homotopy equivalent to ${S}^{1}$, but that is misleading because the edges are glued together. The linked demonstration makes it clear that $U$ is the union of two circular bands, whose intersection is a filled square (and so is contractible). It therefore follows from Lemma 20.1 that ${H}_{0}(U)=\mathbb{Z}$ and ${H}_{1}(U)=\mathbb{Z}\oplus \mathbb{Z}$ and ${H}_{k}(U)=0$ for $k>1$. More specifically, the two bands are thickenings of the loops $v$ and $w$, so ${H}_{1}(U)=\mathbb{Z}\{[v],[w]\}$. All the relevant spaces are connected, so ${H}_{0}(T)=\mathbb{Z}$ and we have a truncated Mayer-Vietoris sequence as in Proposition 15.2. After filling in the known groups, we see that the tail end of this sequence is as follows:

$$0\to {H}_{2}(T)\stackrel{\mathit{\delta}}{\to}\mathbb{Z}.[u]\stackrel{{i}_{*}}{\to}\mathbb{Z}\{[v],[w]\}\stackrel{{k}_{*}}{\to}{H}_{1}(T)\to 0.$$ |

In $U$, we can deform $u$ outwards to the edge of the square (which does not change the homology class). It then becomes equal to the join $v*w*\overline{v}*\overline{w}$. As discussed in Proposition 10.29, this is homologous to $v+w-v-w=0$. This proves that ${i}_{*}=0$, so $\mathrm{ker}({i}_{*})=\mathbb{Z}.[u]$ and $\mathrm{img}({i}_{*})=0$. As the sequence is exact, it follows that $\mathrm{img}(\delta )=\mathbb{Z}.[u]$ and $\mathrm{ker}({k}_{*})=0$, so the maps $\delta $ and ${k}_{*}$ are isomorphisms, so ${H}_{1}(T)\simeq {\mathbb{Z}}^{2}$ and ${H}_{2}(T)\simeq \mathbb{Z}$. For $k\ge 3$ we have an exact sequence

$$0={H}_{k}(U)\oplus {H}_{k}(V)\to {H}_{k}(X)\stackrel{\mathit{\delta}}{\to}{H}_{k-1}(U\cap V)=0,$$ |

which shows that ${H}_{k}(T)=0$. ∎

As an alternative, we could write the torus as the union of the following open sets:

Here $U$ and $V$ are both homotopy equivalent to ${S}^{1}$, whereas $U\cap V$ is homotopy equivalent to the union of two disjoint copies of ${S}^{1}$. We leave it to the reader to understand how the resulting Mayer-Vietoris sequence gives the same answer as before. This method can be generalised to calculate the homology of the $d$-dimensional torus ${T}_{d}={({S}^{1})}^{d}={S}^{1}\times \mathrm{\cdots}\times {S}^{1}$. The answer is that ${H}_{k}({T}_{d})\simeq {\mathbb{Z}}^{\left(\genfrac{}{}{0pt}{}{d}{k}\right)}$ for all $k$, but we will not give the details here.

For the real projective plane $\mathbb{R}\mathit{}{P}^{\mathrm{2}}$ we have ${H}_{\mathrm{0}}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{\mathrm{2}}\mathrm{)}\mathrm{=}\mathbb{Z}$ and ${H}_{\mathrm{1}}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{\mathrm{2}}\mathrm{)}\mathrm{=}\mathbb{Z}\mathrm{/}\mathrm{2}$ and ${H}_{k}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{\mathrm{2}}\mathrm{)}\mathrm{=}\mathrm{0}$ for $k\mathrm{\ge}\mathrm{2}$.

Recall (perhaps from the Knots and Surfaces course) that we can form $\mathbb{R}{P}^{2}$ by taking a disc as shown on the left below, and identifying opposite points on the boundary circle. In particular, the two points marked $a$ are the same, and each point on the upper red semicircle is identified with the corresponding point on the lower blue semicircle, so the two paths marked $v$ are the same. We consider $\mathbb{R}{P}^{2}$ as the union of the indicated open sets $U$ and $V$.

It is clear that $V$ is contractible, so ${H}_{0}(V)=\mathbb{Z}$ and ${H}_{k}(V)=0$ for $k\ge 1$. It is also clear that that $U\cap V$ is homotopy equivalent to a circle, so ${H}_{0}(U\cap V)=\mathbb{Z}$ and ${H}_{1}(U\cap V)=\mathbb{Z}.[u]$ and ${H}_{k}(U\cap V)=0$ for $k\ge 2$.

Next, we claim that $U$ is homeomorphic to a Möbius strip, and therefore homotopy equivalent to a circle. One way to see this is to use the following animated diagram:

Alternatively, we can argue using the pictures below. The first one shows $U$ together with two extra edges marked $p$ and $q$. If we cut along these, we get the middle picture. If we flip the top half over and glue it to the bottom half along $v$, we get the third picture, which is the standard gluing diagram for a Möbius strip.

Either way, it follows that ${H}_{0}(U)=\mathbb{Z}$ and ${H}_{1}(U)=\mathbb{Z}.[v]$ and ${H}_{k}(U)=0$ for $k\ge 2$. We next need to understand the map ${i}_{*}:{H}_{1}(U\cap V)\to {H}_{1}(U)$. In $U$, we can push the loop $u$ out to the boundary by a loop homotopy, which does not change the homology class. The deformed loop then covers both copies of $v$, so we have ${i}_{*}([u])=2[v]$. After taking account of the known homology groups, the end of the truncated Mayer-Vietoris sequence looks like this:

As ${i}_{*}([u])=2[v]$ we see that $\mathrm{ker}({i}_{*})=0$ and so exactness forces ${H}_{2}(\mathbb{R}{P}^{2})$ to be zero. Also, the map ${k}_{*}$ is surjective with kernel given by the image of ${i}_{*}$, which is $2\mathbb{Z}.[v]$, so ${H}_{1}(\mathbb{R}{P}^{2})\simeq \mathbb{Z}/2$. For $k\ge 3$ we also have exact sequences

$$0={H}_{k}(U)\oplus {H}_{k}(V)\to {H}_{k}(\mathbb{R}{P}^{2})\to {H}_{k-1}(U\cap V)=0,$$ |

so ${H}_{k}(\mathbb{R}{P}^{2})=0$. As $\mathbb{R}{P}^{2}$ is connected, we have ${H}_{0}(\mathbb{R}{P}^{2})=\mathbb{Z}$. ∎

Now consider a closed surface $X$ presented as in the Knots and Surfaces course, by gluing edges of a polygon according to a surface word. We illustrate this using the standard word $ab{a}^{-1}{b}^{-1}cd{c}^{-1}{d}^{-1}$ for an orientable surface of genus $2$, but the method works for any word satisfying the usual conditions of surface theory. We can calculate ${H}_{*}(X)$ using essentially the same method that we used for the torus. The conclusion is that ${H}_{0}(X)={H}_{2}(X)=\mathbb{Z}$, and ${H}_{k}(X)=0$ for $k\ge 3$, and ${H}_{1}(X)={\mathbb{Z}}^{2g}$, where $g$ is the genus. In particular, in the illustrated case we have ${H}_{1}(X)={\mathbb{Z}}^{4}$. To see this, we use open sets $U$ and $V$ illustrated below.

As with the case of the torus, the space $V$ is contractible and the space $U\cap V$ is homotopy equivalent to a circle, so their homology is easy to understand. The space $U$ apparently has $8$ edges and $8$ marked vertices. However, the edges are glued together in pairs, in such a way that all vertices get glued together; the result is just the space ${W}_{4}$ from Example 20.4. The full space $U$ consists of ${W}_{4}$ with a fringe attached, but that does not affect the homotopy type, so we have ${H}_{*}(U)={H}_{*}({W}_{4})$. This means that ${H}_{0}(U)=\mathbb{Z}$ and ${H}_{1}(U)={\mathbb{Z}}^{4}$ and ${H}_{k}(U)=0$ for $k\ge 2$. If we let $u$ be a loop once around $U\cap V$, then in $U$ we see that $u$ becomes $a*b*\overline{a}*\overline{b}*c*d*\overline{c}*\overline{d}$. This has homology class $[a]+[b]-[a]-[b]+[c]+[d]-[c]-[d]=0$, so the homomorphism ${i}_{*}:{H}_{1}(U\cap V)\to {H}_{1}(U)$ is zero. We now have an exact sequence

$$0\to {H}_{2}(X)\stackrel{\mathit{\delta}}{\to}\mathbb{Z}\stackrel{{i}_{*}=0}{\to}{\mathbb{Z}}^{4}\stackrel{}{\to}{H}_{1}(X)\to 0,$$ |

and the claimed description of ${H}_{*}(X)$ follows easily from this.

The group ${H}_{\mathrm{1}}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{\mathrm{1}}\mathrm{)}$ is isomorphic to $\mathbb{Z}$, but for $n\mathrm{\ge}\mathrm{2}$ the group ${H}_{\mathrm{1}}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{n}\mathrm{)}$ is isomorphic to $\mathbb{Z}\mathrm{/}\mathrm{2}$. Moreover, for $k\mathrm{>}n$ we have ${H}_{k}\mathit{}\mathrm{(}\mathbb{R}\mathit{}{P}^{n}\mathrm{)}\mathrm{=}\mathrm{0}$.

We saw in Example 7.23 that $\mathbb{R}{P}^{1}$ is homeomorphic to ${S}^{1}$, so ${H}_{1}(\mathbb{R}{P}^{1})\simeq \mathbb{Z}$. The case $n=2$ is given by Proposition 20.7. We have an inclusion $i:{S}^{n}\to {S}^{n+1}$ given by $i({x}_{0},\mathrm{\dots},{x}_{n})=({x}_{0},\mathrm{\dots},{x}_{n},0)$. This satisfies $i(-x)=-i(x)$, so it induces an inclusion ${S}^{n}/\{\pm 1\}\to {S}^{n+1}/\{\pm 1\}$, or in other words $\mathbb{R}{P}^{n}\to \mathbb{R}{P}^{n+1}$; we again call this $i$. It will be enough to check that the map ${i}_{*}:{H}_{1}(\mathbb{R}{P}^{n})\to {H}_{1}(\mathbb{R}{P}^{n+1})$ is an isomorphism for $n\ge 2$.

Recall that ${S}^{n+1}=\{x\in {\mathbb{R}}^{n+2}|{\parallel x\parallel}^{2}=1\}$. We can identify ${\mathbb{R}}^{n+2}$ with ${\mathbb{R}}^{n+1}\times \mathbb{R}$; then

$${S}^{n+1}=\{(y,z)\in {\mathbb{R}}^{n+1}\times \mathbb{R}|{\parallel y\parallel}^{2}+{z}^{2}=1\}.$$ |

The subspace where $z=0$ can be identified with ${S}^{n}$. Now let ${U}_{+}$ and ${U}_{-}$ be the subspaces where $z>0$ and $$ respectively, and put $\stackrel{~}{U}={U}_{+}\cup {U}_{-}$. We also put

$$ |

Now let $\pi :{S}^{n+1}\to {S}^{n+1}/\{\pm 1\}=\mathbb{R}{P}^{n+1}$ be the quotient projection (so $\pi (x)=\pi ({x}^{\prime})$ iff ${x}^{\prime}=\pm x$). Put $U=\pi (\stackrel{~}{U})$ and $V=\pi (\stackrel{~}{V})$. These sets are easily seen to be open with respect to the quotient topology on $\mathbb{R}{P}^{n+1}$.

Let ${U}^{\prime}$ be the open ball of radius one in ${\mathbb{R}}^{n+1}$. We have maps

$${U}^{\prime}\stackrel{\mathit{f}}{\to}{U}_{+}\stackrel{\mathit{\pi}}{\to}U,$$ |

where $f(y)=(y,\sqrt{1-{\parallel y\parallel}^{2}})$, and one can check that both of these are homeomorphisms. It follows that $U$ is contractible. The same maps also give homeomorphisms

$${U}^{\prime}\setminus \{0\}\stackrel{\mathit{f}}{\to}{U}_{+}\setminus \{(0,1)\}\stackrel{\mathit{\pi}}{\to}U\cap V.$$ |

It follows that $U\cap V$ is homotopy equivalent to ${S}^{n}$.

Next, let $j:{S}^{n}\to \stackrel{~}{V}$ be the inclusion, and define $r:\stackrel{~}{V}\to {S}^{n}$ by $r(y,z)=y/\parallel y\parallel $ (which is valid because $y\ne 0$ when $(y,z)\in \stackrel{~}{V}$). This satisfies $r\circ j=\mathrm{id}:{S}^{n}\to {S}^{n}$. We also define $h:[0,1]\times \stackrel{~}{V}\to \stackrel{~}{V}$ by

$$h(t,(y,z))=(y,tz)/\sqrt{{\parallel y\parallel}^{2}+{t}^{2}{z}^{2}};$$ |

this gives a homotopy between $j\circ r$ and the identity. These maps satisfy $j(-x)=-j(x)$ and $r(-x)=-r(x)$ and $h(t,-x)=-h(t,x)$ so they induce maps $j:\mathbb{R}{P}^{n}\to V$ and $r:V\to \mathbb{R}{P}^{n}$ and $h:[0,1]\times V\to V$. Using this we see that $j$ is a homotopy equivalence, so ${H}_{1}(V)={H}_{1}(\mathbb{R}{P}^{n})$.

The spaces $U$, $V$, $U\cap V$ and $U\cup V=\mathbb{R}{P}^{n+1}$ are all path connected, so we have a truncated Mayer-Vietoris sequence

$${H}_{1}(U\cap V)\to {H}_{1}(U)\oplus {H}_{1}(V)\to {H}_{1}(\mathbb{R}{P}^{n+1})\to 0.$$ |

Here $U\cap V$ is homotopy equivalent to ${S}^{n}$ with $n\ge 2$ so ${H}_{1}(U\cap V)=0$. The space $U$ is contractible so ${H}_{1}(U)=0$. The space $V$ is homotopy equivalent to $\mathbb{R}{P}^{n}$, so we can assume inductively that ${H}_{1}(V)=\mathbb{Z}/2$. Exactness of the sequence now implies that ${H}_{1}(\mathbb{R}{P}^{n+1})=\mathbb{Z}/2$ as required.

Now suppose that $k>n+1$. We have a Mayer-Vietoris sequence

$${H}_{k}(U)\oplus {H}_{k}(V)\to {H}_{k}(\mathbb{R}{P}^{n+1})\to {H}_{k-1}(U\cap V),$$ |

or equivalently

$${H}_{k}(\mathbb{R}{P}^{n})\to {H}_{k}(\mathbb{R}{P}^{n+1})\to {H}_{k-1}({S}^{n}).$$ |

As $k>n+1$ we have ${H}_{k-1}({S}^{n})=0$. Also, we can assume inductively that ${H}_{j}(\mathbb{R}{P}^{n})=0$ for all $j>n$, so in particular ${H}_{k}(\mathbb{R}{P}^{n})=0$. It follows by exactness that ${H}_{k}(\mathbb{R}{P}^{n+1})=0$ as claimed. ∎

The full story is as follows. For $m\ge 0$ we have

${H}_{k}(\mathbb{R}{P}^{2m+1})$ | $$ | ||

${H}_{k}(\mathbb{R}{P}^{2m+2})$ | $$ |

Equivalently, let $P(n)$ be the following chain complex:

$$\mathrm{\cdots}\stackrel{}{\leftarrow}0\stackrel{}{\leftarrow}0\stackrel{}{\leftarrow}\mathbb{Z}\stackrel{0}{\leftarrow}\mathbb{Z}\stackrel{2}{\leftarrow}\mathbb{Z}\stackrel{0}{\leftarrow}\mathbb{Z}\stackrel{2}{\leftarrow}\mathrm{\cdots}\stackrel{}{\leftarrow}\mathbb{Z}\stackrel{}{\leftarrow}0\stackrel{}{\leftarrow}0\stackrel{}{\leftarrow}\mathrm{\cdots}.$$ |

There are copies of $\mathbb{Z}$ in degrees $0$ to $n$ inclusive, and the differentials alternate between zero and multiplication by $2$. Then ${H}_{*}(\mathbb{R}{P}^{n})\simeq {H}_{*}(P(n))$ for all $n$. This can be proved using the same Mayer-Vietoris sequence as mentioned above, but some extra work is needed to determine the maps in that sequence.