Codex @codex  0

If the one-point compactification $X^{+}$ is Hausdorff and has a basis of contractible neighbourhoods at the added point, thenThe Excision theorem identifies $H^{\ast }(X,X\setminus K)$ with $H^{\ast }(X^{+},X^{+}\setminus K)$. The contractible neighbourhoods form a cofinal family of the complements of compact $K$, and their pair long exact sequences identify each relative group naturally with reduced cohomology. Passing to the direct limit gives the comparison. Local contractibility matters for singular cohomology; the Hawaiian earring illustrates its failure.

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

Define compactly supported cohomology byFor $K\subseteq L$, the transition map is induced by the identity map of pairs $(X,X\setminus L)\to (X,X\setminus K)$. Equivalently, take the cochain complex of singular cochains that vanish on every chain contained in the complement of some compact set. Directed unions are exact, giving the same definition.

For $\mathbb{R}$, the intervals $[-a,a]$, $a>0$, are cofinal among compact subsets. The complement has two contractible components. The long exact sequence in relative cohomology contains the diagonal map $\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$, so its cokernel is $H^1(\mathbb{R},\mathbb{R}\setminus [-a,a])\cong \mathbb{Z}$, and all the other relative groups vanish. Enlarging the interval preserves the generator given by the difference of the two ends. Therefore

For the compact-support comparison with a one-point compactification, write $Y=X^{+}$. The assumed Hausdorff one-point compactification is compact; a compact subset $K\subseteq X$ is closed in $Y$. The Excision theorem removes $\{\infty \}$ from the pair $(Y,Y\setminus K)$, because its closure lies inside the open second member. ThusComplements of compact subsets of $X$ are exactly the open neighbourhoods of $\infty$ in $Y$. The hypothesis supplies a cofinal family of contractible such neighbourhoods $U$. For every one, the long exact sequence of the pair identifiesIn degree zero, this is the kernel of evaluation on the component of $\infty$, identified with reduced cohomology by subtracting the constant value there. In degree one the map $H^0(Y)\to H^0(U)\cong \mathbb{Z}$ is surjective; in higher degrees the positive cohomology of $U$ vanishes. These identifications are natural for inclusions of contractible neighbourhoods. Passing to the direct limit proves

For the specified disjoint union of lines, a compact subset meets only finitely many components and is bounded in each. Finite unions $F\times [-a,a]$, with $F\subset \mathbb{Z}$ finite, are cofinal. Applying the preceding relative calculation componentwise givesThe one-point compactification $W$ of this space is the Hawaiian earring: each line becomes a circle by adding the common point $\infty$, and every neighbourhood of $\infty$ contains all but finitely many whole circles. On the remaining finitely many circles it contains neighbourhoods of the common point. This describes exactly the shrinking-circle topology. In particular $W$ is not locally contractible at $\infty$: every such neighbourhood contains a whole circle, whose generator remains nontrivial under the retraction $W\to S^1$ that collapses all the other circles.

For integral singular cohomology, the comparison does not hold. Here is a degree-two obstruction that takes account of the shrinking-circle topology. The standard rational summand in Hawaiian earring homology theorem gives a direct summand $\mathbb{Q}$ in $H_1(W;\mathbb{Z})$. The universal coefficient theorem for cohomology injectsThe summand $\mathbb{Q}$ therefore contributes the nonzero Ext of the rationals with integer coefficients.

For completeness, this last algebraic assertion has an explicit proof. Present $\mathbb{Q}$ using generators $a_n=1/n!$ and relations $a_n-(n+1)a_{n+1}=0$, $n\ge 1$. The corresponding free resolution shows that ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})$ is the cokernel ofThe constant sequence $(1,1,\dots )$ is not in the image. Otherwise iteration would giveFor large $N$, the factorial sum exceeds $|u_1|$ but is less than $(N+1)!-|u_1|$, making that congruence impossible. Thus the cokernel is nonzero. It follows thatwhich proves the failure of the claimed isomorphism. The ingredient concerning the Hawaiian earring is its singular-homology structure theorem, not the homology of an infinite CW complex wedge of circles; these topologies differ.

A singular $n$-cochain with coefficients in an abelian group $A$ is a homomorphism from the singular chain group $C_n(X;\mathbb{Z})$ to $A$. It is specified by its values on all singular simplices, without a finite-support restriction. Its coboundary evaluates on the alternating sum of the faces. The resulting cochain complex computes singular cohomology.