Singular homology of a quotient of $\mathbb{C}P^n$ by an order $2$ action. Let $\mathbb{C}P^n$ be complex projective $n$-space. Let $f:\mathbb{C}P^n\rightarrow \mathbb{C}P^n$ be the map given by $[x_0,\dots,x_n]\rightarrow [-x_0,\dots,x_n]$.
Let $X:=\mathbb{C}P^n/\{1,f\}$ be the orbit space by this action. I would like to complete the singular homology groups of $X$. My conjecture is this: $H_i(X)=\mathbb{Z}/2\mathbb{Z}$ iff $i\leq n$ and $i$ is even and it is zero otherwise.
I have attempted the following methods to do so. The second method appears to have a higher chance of success.
First, I tried Mayer-Vietoris by looking at the quotients $S^{2n+1}\rightarrow \mathbb{C}P^n\rightarrow X$ and trying to take the hemispheres from $S^{2n+1}$ down to $X$ as the choice of cover. My only complaint here is that I don't understand what the quotient ends up doing to this choice of cover very well.
Second, I tried using cellular homology directly. We know $\mathbb{C}P^n$ has a CW-structure given by $e^0\cup\dots\cup e^{2n}$ i.e. we glue a cell in each even dimension up to $2n$. The gluing map is the quotient map $S^{2k+1}\rightarrow \mathbb{C}P^{k}$ at each step. Now for $X$, I would like to claim $X$ has a CW-complex structure (but this need not hold in general). In the dummy case of $n=1$, I know that $X=\mathbb{R}P^2$. But I cannot see it beyond this situation...
A hint would be appreciated as opposed to a full solution!
 A: The short version is that

$H_\ast(X)\cong \begin{cases}\mathbb{Z} & 0\leq \ast\leq 2n, \ast \text{ even}\\ 0 & \text{otherwise} \end{cases}$.

Here's a proof.  We'll start by setting up notation.  I'm going to think of $\mathbb{C}P^{n}$ as the mapping cone, $C(\pi)$ of the Hopf map $\pi:S^{2n-1}\rightarrow \mathbb{C}P^{n-1}$.  That is $\mathbb{C}P^n$ is homeomorphic to $C(\pi):=S^{2n-1}\times [0,1]/\sim$ where $\sim$ identifies $S^{2n-1}\times \{0\}$ to a single point, and it identifies points in $S^{2n-1}\times \{1\}$ if they are in the same Hopf orbit.  If you think about it this is nothing but the usual cell stucture on $\mathbb{C}P^n$:  The subspace consisting of points in $S^{2n-1}\times [0,1)/\sim$ is a disk $D^{2n}$, which is being glued to $\mathbb{C}P^{n-1} = S^{2n-1}\times \{1\}/\sim$ via the Hopf map.
An explicit homeomorphism is given by $\phi:\mathbb{C}P^n\rightarrow C(\pi)$ defined by $\phi([z_0:z_1:...:z_n]) = \left(\frac{|z_0|}{\sqrt{1-|z_0|}^2}\left(\frac{z_1}{z_0},...., \frac{z_n}{z_0}\right), 1-|z_0|\right)$.  Note that when $|z_0| = 1$, the first coodinate doesn't really make sense, but all such points get identified in the quotient.  Likewise, when $|z_0| = 0$, the first coordinate doesn't make sense, but it still does as  point in $\mathbb{C}P^{n-1}$.
What does $f$ look like in this description?  Well, away from the points with $|z_0| = 0,1$, it's just the antipodal map.  But at the single point where $|z_0| = 1$, it fixes that point, and it also fixes the $\mathbb{C}P^{n-1}$ worth of points where $|z_0| = 0$.
Thus, we may describe $C(\pi)/f$ as $\mathbb{R}P^{2n-1}\times [0,1]/\sim$, where $\sim$ works the same way.
Said another way, $\mathbb{C}P^n/f$ is $C(\rho)$ where $\rho:\mathbb{R}P^{2n-1}\rightarrow \mathbb{C}P^{n-1}$ is real-projective Hopf map.
We are now ready to compute the topology, using Mayer-Vietoris.  Fix $\epsilon$ to be a tiny real number and let $U\subseteq C(\rho)$ with $U = \{([x,t]\in C(\rho): t\in [0, 1/2 + \epsilon)\}$ and also let $V = \{([x,t]\in C(\rho): t\in (1/2-\epsilon, 1]\}$.
The space $U$ is the open cone on $\mathbb{R}P^{2n-1}$, so clearly deformation retracts to a point.  The space $V$ is the mapping cylinder of $\rho$, so clearly deformation retracts to $\mathbb{C}P^{n-1}$.
The subspace $U\cap V$ deformation retracts to $\mathbb{R}P^{2n-1}\times \{1/2\}$.
The last bit of info we need is what the inclusion map $U\cap V\rightarrow V$ does on homology.  Of course, in dimension $0$, it's an isomorphism.  Otherwise, it's the trivial map:  $U\cap V$ only has homology in odd degrees while $V$ only has homology in even degrees.  We'll see later that we need a bit more: what the inclusion map does on homology with $\mathbb{Z}/2\mathbb{Z}$ coefficients.
Proposition:  Consider the inclusion map $i:U\cap V\rightarrow V$.  Then the induced map $$i_\ast:H_{2k}(U\cap V;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}\cong H_{2k}(V; \mathbb{Z}/2\mathbb{Z})$$ is an isomorphism.
Proof:  Up to homotopy, the map $i$, followed by the deformation retract from $V$ to $\mathbb{C}P^{n-1}$, is nothing but $\rho$.  Thus, we are really interested in determining $\rho_\ast$.
To that end, consider $\mathbb{R}P^{2k}\subseteq \mathbb{R}P^{2n-1}\times \{1/2\}$, where we think of $\mathbb{R}P^{2n-1}$ as classes $\langle z_0:...:z_n\rangle\in S^{2n-1}/a$, where $a:S^{2n-1}\rightarrow S^{2n-1}$ is the antipodal map, and where we think of $\mathbb{R}P^{2k}$ as the subset set $\langle 0:..: t: z_{n-k+1}:...:z_n\rangle$ with $t$ real.  (So, given a point $(z_0,...,z_n)\in S^{2n+1}\subseteq \mathbb{C}P^{n+1}$, I'm using the notation $\langle z_0:...:z_n\rangle$ to denote its image in $\mathbb{R}P^{2n+1}$, and $[z_0:...:z_n]$ for the image in $\mathbb{C}P^{n}$.)
It is well known that $[\mathbb{R}P^{2k}]$ generates $H_{2k}(\mathbb{R}P^{2n-1};\mathbb{Z}/2\mathbb{Z})$.  Thus, we must determine $\rho([\mathbb{R}P^{2k}]) = [\rho(\mathbb{R})P^{2k})]$.
Now, $\rho|_{\mathbb{R}P^{2k}}\mathbb{R}P^{2k}\rightarrow \mathbb{C}P^{n-1}$ has image in $\mathbb{C}P^k\subseteq \mathbb{C}P^{n-1}$, which has the same dimension as $\mathbb{R}P^{2k}$.  So we can just compute its mod $2$ degree in the usual sense.
If we consider a point $[\cos(\theta):\sin(\theta)z_{n-k+1}:...:\sin(\theta)z_n]$, then it's not too hard to verify that this gives a regular point whenever both trig functions are non-vanishing.  Such a point has preimage consisting of precisely one point in $\mathbb{R}P^{2k}$:  the point $\langle 0:...:t:\sin(\theta)z_{n-k+1}:...:\sin(\theta)z_n\rangle$ with $t = \pm \cos(\theta)$.  Thus, the mod $2$ degree of the map is $1$, establishing the proposition.  $\square$
Armed with all this info, we can apply Mayer-Vietoris.  Beginning with integer coefficients, all the map $H_k(U\cap V)\rightarrow H_k(U)\oplus H_k(V)$ are trivial (except when $k =0$, of course), so we obtain short exact sequences of the form $$0\rightarrow H_k(U)\oplus H_k(V)\rightarrow H_k(X)\rightarrow H_{k-1}(U\cap V)\rightarrow 0.$$
When $k$ is odd, $H_k(V) = H_{k-1}(U\cap V) = 0$, which implies $H_k(X) = 0$.
Wen $k = 2n$, we instead get $$0\rightarrow 0\rightarrow H_k(X)\rightarrow  \mathbb{Z}\rightarrow 0,$$ so $H_k(X)\cong \mathbb{Z}$.
When $k< 2n$ is even, $H_k(V)\cong \mathbb{Z}$ and $H_{k-1}(U\cap V) \cong \mathbb{Z}/2\mathbb{Z}$, so we get an exact sequence of the form $$0\rightarrow \mathbb{Z}\rightarrow H_k(X)\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 0.$$
This tells us that $H_k(X)\cong \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ or $H_k(X)\cong \mathbb{Z}$, but it doesn't pin down which option occurs.
To answer this, we'll rerun Mayer-Vietories with $\mathbb{Z}/2\mathbb{Z}$ coefficients.  (I'll use $R$ instead of $\mathbb{Z}/2\mathbb{Z}$ just to save on typing.)   It's still true that when $k$ is odd, that $H_k(U;R)\oplus H_k(V;R) = 0$.  Thus the Mayer-Vietoris sequence, beginning with $0 = H_{2k+1}(U;R)\oplus H_{2k+1}(V;R)$, breaks into longer sequences of the form $$ 0\rightarrow H_{2k+1}(X;R)\rightarrow H_{2k}(U\cap V;R)\rightarrow H_{2k}(V;R)\rightarrow H_{2k}(X;R)\rightarrow H_{2k-1}(U\cap V;R)\rightarrow 0 .$$
This is $$0\rightarrow H_{2k+1}(X;R)\rightarrow R\rightarrow R\rightarrow H_{2k}(X;R)\rightarrow R\rightarrow 0.$$
The map $R\rightarrow R$ is the one from the proposition, so we know it's an isomorphism.  Thus, we conclude $H_{2k+1}(X;R) = 0$ and $H_{2k}(X;R) \cong R$.
Since we already know that $H_{2k}(X;\mathbb{Z}) \cong \mathbb{Z}$ or $\mathbb{Z}\oplus R$, universal coefficients now tells us that it must be the first option.
