homology groups of a product-quotient of spheres In many textbooks the integral homology groups of an $n$-dimensional sphere $S^n$ and of its quotient, the real projective space $\mathbb{R}\mathbb{P}^n=S^n/\{\pm 1\}$ are explained. As further examples, I want to work out the integral homology groups of the product-quotient
$$X_{m,n}:= (S^m\times S^n)/\{\pm 1\}$$,
which is doubly covered by $S^m\times S^n$ and doubly covers $\mathbb{R}\mathbb{P}^m \times \mathbb{R}\mathbb{P}^n$. But, for general $m$ and $n$, I encountered difficulty in computing especially the groups of intermediate degrees (see below). I have tried some cases:

*

*If $n=1$ and $m\geq 2$, then the projection $X_{m,1}\rightarrow S^1/\{\pm 1\}$ exhibits $X_{m,1}$ as a sphere bundle over a circle whose geometric monodromy acts a fiber by $-1$. By the Mayer-Vietoris sequence, I could compute the homology groups as
$$\text{if $m$ is even,}\qquad  H_i(X_{m,1},\mathbb{Z})=
\begin{cases}
&\mathbb{Z}\quad (i=0,1)\\
&\mathbb{Z}/2 \ (i=m)\\
&0\qquad \text{otherwise.}
\end{cases}$$
$$\text{if $m$ is odd and $>1$,}\quad  H_i(X_{m,1},\mathbb{Z})=
\begin{cases}
&\mathbb{Z}\quad (i=0,1,m,m+1)\\
&0\qquad \text{otherwise.}
\end{cases}$$
This case looks complete.


*For general $m,n$, using Kuenneth formula and [Hatcher's algebraic topology, Proposition 3G.1] I see the Betti numbers as the number of $(-1)$-invariant cohomology classes on the product.
In particular, I see the rational homology groups.


*For $m=n=2$: By 2. above we see $H_0=H_4=\mathbb{Z}$ and $H_1=\mathbb{Z}/2$ since
its fundamental group is of order 2. The remaining degree $2,3$ can be obtained using Poincare duality and the universal coefficient formula as $H_2= \mathbb{Z}/2$ and $H_3=0$. This case can be understood.


*For $m=2,n=4$: The same method as above computes $H_0,H_1,H_4,H_5,H_6$ but the remaining two groups (both torsion) seems not to be reachable by my ad hoc method.
Question: How can we compute the full integral homology of $X_{m,n}$ ?
Maybe a good CW-structure should be introduced, but I couldn't figure out. Any help is appreciated. Also, please point and correct my computations above if there are missing stuffs. Thank you in advance.
 A: Let $t$ denote the antipodal map on $S^n$.  There is a simple CW structure for $S^n$ compatible with $t$, which we define inductively by $S^0=\{*,*t\}$, for a point $*$, and $$S^n=S^{n-1}\cup_\phi e_n\cup_\psi e_nt,$$ where $e_n$ is an $n$-cell and $\phi,\psi$ are identifications of the boundaries of $e_n$ and $e_nt$ with $S^{n-1}$.
Intuitively, we regard each sphere as the equator of the next, and glue on a couple of hemispheres.
The boundary maps $d$ are given by $$de_r=e_{r-1}(1+(-1)^rt),\qquad de_0=0.$$
To understand the sign $(-1)^r$ here, note if $r$ is odd then $t$ preserves the orientation of $S^{r-2}$, so the boundary of $e_r$ must be the difference $e_{r-1}-e_{r-1}t$ in order to be closed.  Conversely if $r$ is even, then $t$ reverses the orientation of $S^{r-2}$, so the boundary of $e_r$ must be the sum $e_{r-1}+e_{r-1}t$ in order to be closed.
We may take the tensor product of the resulting chain complexes for $S^n$ and $S^m$, to obtain a chain complex for $S^n\times S^m$.
Here the degree $r$ chain group is generated by \begin{eqnarray*}e_a\otimes e_{r-a},\\ e_at\otimes e_{r-a},\\ e_a\otimes e_{r-a}t,\\ e_at\otimes e_{r-a}t.\end{eqnarray*} for $a=0,\cdots, r|a\leq n, r-a\leq m$.  The differential is given by \begin{eqnarray*}d(e_i\otimes e_j)&=&e_{i-1} \otimes e_j\\ + &(-1)^i& e_{i-1}t \otimes e_j\\+&(-1)^i& e_{i} \otimes e_{j-1}\\&(-1)^{i+j}& e_{i} \otimes e_{j-1}t.
\end{eqnarray*}
That is we use Leibniz's rule for tensor products:$$d(a\otimes b)=da\otimes b+ (-1)^{{\rm deg}(a)} a\otimes db.$$
Finally, we obtain a chain complex for $$X=(S^n\times S^m)/\langle t\rangle,$$
by taking our chain complex and identifying $$e_i\otimes e_j\sim e_it\otimes e_jt,\qquad e_i\otimes e_jt\sim e_it\otimes e_j.$$
We may write the boundary maps $d$ as matrices, with respect to the basis $\ldots e_a\otimes e_b, e_at\otimes e_b, e_{a+1}\otimes e_{b-1},\ldots$.
The homology groups of $X$ may then be computed as an exercise in linear algebra.  For example, if $0<<m<<n$ and $n,m$ both odd I get that the non-zero homology groups are:
\begin{eqnarray*} H_0(X)\cong H_{n+m}(X)&\cong& \mathbb{Z},\\
H_m(X)\cong H_n(X)&\cong&\mathbb{Z},\\
H_{2i-1}(X)\cong H_{n+m-2i}&\cong& \mathbb{Z}/2\mathbb{Z}, \qquad{\rm for} \,\,i=1,\cdots,(m-1)/2.\end{eqnarray*}
I suggest that you do this calculation yourself, as it is quite fiddly and I may have made an error.  However the above answer has the symmetries one would expect.  If you require help, I can give you the matrices.
