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For a nonnegative integer $d$ and an abelian group $A$,The one-point compactification is the sphere $S^d$, and it is locally contractible at the added point. Apply the compact-support comparison with a one-point compactification and the reduced cohomology of a sphere. In dimension zero the compactification is $S^0$, giving the same formula.

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.

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.

The quaternionic projective space ${\mathbb{HP}}^n$ is the space of one-dimensional right quaternion subspaces of ${\mathbb{H}}^{n+1}$. Equivalently it is the quotient of the unit sphere $S^{4n+3}$ by simultaneous right multiplication by unit quaternions. Its coordinate filtration has one open cell ${\mathbb{H}}^k\cong {\mathbb{R}}^{4k}$ in each dimension $4k$, for $0\le k\le n$. Hence its cellular cohomology is $\mathbb{Z}$ in those dimensions and zero otherwise.

Let $\mathcal{H}$ be the quaternionic tautological line bundle. Its unit sphere bundle is $S^{4n+3}\to {\mathbb{HP}}^n$, with fibre $S^3$. The Gysin sequence of a sphere bundle shows that multiplication by its Euler class $e\in H^4$ is an isomorphism from $H^{4k}$ to $H^{4k+4}$ for $k<n$. Choose the generator $u=-e$. Its powers generate every nonzero positive degree, giving the cohomology ring of quaternionic projective spaceThis also accounts for $n=0$.

First take $n=2$. Under the coordinate inclusion $i:{\mathbb{CP}}^2\to {\mathbb{HP}}^2$, the pulled-back quaternionic line is the quaternionic extension of the complex tautological line $L$. As a complex rank-two bundle it is $L\oplus \overline{L}$: a transition scalar $c\in \mathbb{C}$ acts on the two complex coordinates of a quaternion by $c$ and $\overline{c}$. Put $x=c_1(L^{\ast })$, the degree-two generator of the cohomology ring of complex projective space. The Whitney sum formula for Chern classes givesHere the Euler class of a complex vector bundle is its top Chern class, using the complex orientation. In particular this degree-four pullback has coefficient one; it is not a multiple of larger absolute value. The compatible tautological bundles on the projective filtrations give the same equality for every $n$, and multiplicativity then determines the whole ring map:It is zero whenever $2k>n$. These facts are the complex inclusion into quaternionic projective space.

For an odd prime $p$, the Steenrod reduced powers are natural stable cohomology operationsThey satisfy $P^0=1$, the Cartan formula, $P^{i}a=0$ when $2i>|a|$, and $P^{i}a=a^p$ when $|a|=2i$. In particular, on ${\mathbb{CP}}^{\infty }$ one has $P^{1}x=x^p$ and $P^{i}x=0$ for $i>1$. The Cartan formula and the binomial theorem givePass to the infinite projective spaces, where $i^{\ast }:H^{\ast }({\mathbb{HP}}^{\infty };{\mathbb{F}}_p)\to H^{\ast }({\mathbb{CP}}^{\infty };{\mathbb{F}}_p)$, $u\mapsto x^2$, is injective. The equality just obtained determines the Steenrod powers on quaternionic projective space; restricting to the finite spaces giveswith coefficients modulo $p$ and powers above $n$ set to zero. In particular $P^{1}u=2u^{(p+1)/2}$, $P^{2}u=u^p$, and $P^{i}u=0$ for $i>2$. The infinite-space argument matters: the finite inclusion cannot detect those degrees for which $x^{2k}=0$.

Finally put $Q={\mathbb{HP}}^{15}/{\mathbb{HP}}^{12}$ and $S={\Sigma }^{48}{\mathbb{HP}}^3$. Both have reduced cohomology $\mathbb{Z}$ in degrees $52,56,60$ and zero otherwise. Choose integral generators $a_{13},a_{14},a_{15}$ for $Q$ whose pullbacks under the quotient map are $u^{13},u^{14},u^{15}$, and suspended integral generators $b_1,b_2,b_3$ for $S$ from $u,u^2,u^3$.

Use $p=5$. Naturality for the quotient and the formula above giveStability under the suspension isomorphism instead givesAny homotopy equivalence $h:Q\to S$ would induce isomorphisms on the rank-one integral groups, so $h^{\ast }{b}_1=\epsilon a_{13}$ and $h^{\ast }{b}_3=\delta a_{15}$ with $\epsilon ,\delta \in \{1,-1\}$. Reducing modulo five and commuting $h^{\ast }$ with $P^1$ would require $\epsilon =2\delta$ in ${\mathbb{F}}_5$. Neither $1$ nor $-1$ equals $2$ or $-2$ modulo five. ThereforeThe essential point is that integral generator signs constrain Steenrod comparisons. Arbitrary changes of basis over ${\mathbb{F}}_5$ could rescale these two nonzero coefficients into agreement; a genuine equivalence must also preserve the integral lattices, where only the two signs are available.

A cohomology operation is stable if it commutes with the suspension isomorphism on reduced cohomology. Ordinary cup products become zero on a suspension, but stable operations can still connect its nonzero classes. The Steenrod squares and Steenrod reduced powers are examples.