Homology of countable-dimensional real projective space

This article describes the value (and the process used to compute it) of some homotopy invariant(s) for a topological space or family of topological spaces. The invariant is homology group and the topological space/family is countable-dimensional real projective space
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Statement

Over the integers

The homology groups with coefficients in the ring of integers $\mathbb{Z}$ are given as follows:

$H_p({\mathbb{P}}^{\mathrm{\infty }}(\mathbb{R});\mathbb{Z})=\{\begin{array}{cc}\mathbb{Z}/2\mathbb{Z},& p=1,3,5,\dots \\ 0,& p=2,4,6,\dots \\ \mathbb{Z},& p=0\\ \end{array}$

Over an abelian group or module $M$

The homology groups with coefficients in a module $M$ over a ring $R$ are given by:

$H_p({\mathbb{P}}^{\mathrm{\infty }}(\mathbb{R});M)=\{\begin{array}{cc}M/2M,& p=1,3,5,\dots \\ T,& p=2,4,6,\dots \\ M,& p=0\\ \end{array}$

where $T$ is the 2-torsion submodule of $M$, i.e., the submodule of $M$ comprising elements whose double is zero.

In particular, we see the following cases:

Case on $R$ or $M$	Conclusion about odd-indexed homology groups, i.e., $H_p,p=1,3,5,\dots$	Conclusion about even-indexed homology groups, i.e., $H_p,p=2,4,6,\dots$
$M$ is uniquely 2-divisible, i.e., every element of $M$ has a unique half. This includes the case that $M$ is a field of characteristic not 2.	all zero groups	all zero groups
$M$ is 2-torsion-free, i.e., no nonzero element of $M$ doubles to zero	unclear	all zero groups
$M$ is 2-divisible, but not necessarily uniquely so, e.g., $M=\mathbb{Q}/\mathbb{Z}$	all zero groups	unclear
$M=\mathbb{Z}/2^n\mathbb{Z}$, $n$ any natural number	all isomorphic to $\mathbb{Z}/2\mathbb{Z}$	all isomorphic to $\mathbb{Z}/2\mathbb{Z}$
$M$ is a finite abelian group	all isomorphic to $(\mathbb{Z}/2\mathbb{Z}{)}^r$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$	all isomorphic to $(\mathbb{Z}/2\mathbb{Z}{)}^r$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$

Note that the third case, where $M$ is 2-divisible but not necessarily uniquely so, cannot arise if $M=R$ and it is a unital ring. So when taking coefficients over a unital ring, there is no need to distinguish between 2-divisibility and unique 2-divisibility.

Proof

The chain complex arising from a CW structure is as follows:

$\dots \stackrel{\cdot 0}{\to }\mathbb{Z}\stackrel{\cdot 2}{\to }\mathbb{Z}\stackrel{\cdot 0}{\to }\mathbb{Z}\to \dots \stackrel{\cdot 2}{\to }\mathbb{Z}\stackrel{\cdot 0}{\to }\mathbb{Z}$

where the subscript for the last written entry is $0$, and hence the multiplication by 2 maps arise from even to odd subscripts and the multiplication by zero maps arise from odd to even subscripts.

Homology computation over integers

Case for $p$	Fragment of relevance in chain complex ($(p+1{)}^{th}$ to $p^{th}$ to $(p-1{)}^{th}$)	Cycle group (kernel from $p^{th}$ to $(p-1{)}^{th}$	Boundary group (image from $(p+1{)}^{th}$ group to $p^{th}$ group	Homology group = cycle group/boundary group
0	$\mathbb{Z}\stackrel{\cdot 0}{\to }\mathbb{Z}\stackrel{0}{\to }0$	$\mathbb{Z}$	0	$\mathbb{Z}$
odd	$\mathbb{Z}\stackrel{\cdot 2}{\to }\mathbb{Z}\stackrel{\cdot 0}{\to }\mathbb{Z}$	$\mathbb{Z}$	$2\mathbb{Z}$	$\mathbb{Z}/2\mathbb{Z}$
even, positive	$\mathbb{Z}\stackrel{\cdot 0}{\to }\mathbb{Z}\stackrel{\cdot 2}{\to }\mathbb{Z}$	0	0	0

Homology computation over an abelian group or module $M$

The chain complex remains the same, but each $\mathbb{Z}$ is replaced by $M$.

Denote by $T$ the 2-torsion submodule of $M$ and by $M/2M$ the quotient of $M$ by the submodule $2M$ comprising the doubles of elements.

Case for $p$	Fragment of relevance in chain complex ($(p+1{)}^{th}$ to $p^{th}$ to $(p-1{)}^{th}$)	Cycle group (kernel from $p^{th}$ to $(p-1{)}^{th}$	Boundary group (image from $(p+1{)}^{th}$ group to $p^{th}$ group	Homology group = cycle group/boundary group
0	$M\stackrel{\cdot 0}{\to }M\stackrel{0}{\to }0$	$M$	0	$M$
odd	$M\stackrel{\cdot 2}{\to }M\stackrel{\cdot 0}{\to }\mathbb{Z}$	$M$	$2M$	$M/2M$
even, positive	$M\stackrel{\cdot 0}{\to }M\stackrel{\cdot 2}{\to }M$	$T$	0	$T$