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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.

For the classical subgroup-covering construction, assume $X$ is path-connected, locally path-connected, and semilocally simply connected. These are the usual hypotheses underlying the classification of connected covering spaces; an arbitrary space, as literally allowed by the wording, need not admit every subgroup covering.

Construct the path-class construction of a subgroup covering directly. Consider paths $\alpha$ starting at $x_0$, and identify $\alpha ,\beta$ when they have the same endpoint and the loop class $[\alpha {\beta }^{-1}]$ belongs to $H$. The subgroup properties make this an equivalence relation. Let $Y$ be its set of classes and define $p([\alpha ])=\alpha (1)$.

Choose a path-connected open neighbourhood $U$ of $\alpha (1)$ whose loops become null-homotopic in $X$. Such neighbourhoods form a local basis under the stated local hypotheses. Define a sheetFor each point of $U$ the resulting class is independent of the chosen path $\eta$, since the difference is a loop in $U$, conjugated by $\alpha$, and is trivial in ${\pi }_1(X,x_0)$. Thus $p$ maps this sheet bijectively onto $U$. Give $Y$ the topology generated by these sheets. Restrictions to smaller such neighbourhoods show that the sheets form a basis and that their endpoint maps are homeomorphisms. Two sheets over $U$ that intersect coincide, by concatenating and reversing paths within $U$; otherwise they are disjoint. Therefore $p^{-1}(U)$ is a disjoint union of copies of $U$, and $p$ is a covering map.

The basepoint is the class of the constant path. The path $\alpha$ lifts to its successive prefix classes, giving a path from this basepoint to $[\alpha ]$; hence $Y$ is path-connected. A loop $\gamma$ lifts closed exactly when its final prefix class agrees with the constant class, namely when $[\gamma ]\in H$. Consequently $p_{\ast }({\pi }_1(Y))=H$. Part (i) supplies injectivity, so

The missing local hypothesis is substantive. In the Hawaiian earring, every neighbourhood of the common basepoint contains a whole sufficiently small circle. Its loop is nontrivial in the whole space, since retracting onto that circle preserves it. An evenly covered neighbourhood would lift that loop to a closed loop in one sheet, so the image subgroup of any covering contains a nontrivial element. Therefore the trivial subgroup cannot be realized by a simply connected covering of this space. The intended result is valid under the local covering-space hypotheses above, not for every topological space.

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.

Let $W=(\mathbb{N}\times S^1)/A$, and choose in each circle the point opposite $o$:In the subspace topology from the Euclidean plane, the sequence $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not closed in $X$.

On the other hand, let $D={\bar{f}}^{-1}(B)\subseteq W$. With the quotient map $q$ from part (i),This is closed in $\mathbb{N}\times S^1$: its complement is the open set $\mathbb{N}\times (S^1\setminus \{(0,1)\})$. The definition of the quotient topology, applied to complements, says that $D$ is closed in $W$. A homeomorphism would carry this closed set to the closed set $B$, a contradiction. Therefore $\bar{f}$ is not a homeomorphism.

The obstruction occurs at the common point. The open set $W\setminus D$ contains the collapsed point, but its image $X\setminus B$ contains no neighbourhood of $o$, since $z_n\to o$. The wedge sum topology allows a separate small neighbourhood on each circle; a neighbourhood of $o$ in the Hawaiian earring must contain all sufficiently small whole circles.

Use positive indices $n\ge 1$ and write $p=(0,-1)\in S^1$. SetFor fixed $n$, this function is a homeomorphism from the unit circle to $C_n$: it scales its radius to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the Hawaiian earring $X$ and is surjective onto $X$. It is continuous on each open slice $\{n\}\times S^1$; since these slices form an open cover of $\mathbb{N}\times S^1$, $f$ is continuous on the whole product topology.

Distinct circles $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations areFor $m\ne n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one preimage of $o$, namely $(n,p)$. Therefore the only nonsingleton fibre of $f$ is $A$.

Let $q:\mathbb{N}\times S^1\to (\mathbb{N}\times S^1)/A$ be the quotient map collapsing $A$. The universal property of the quotient topology supplies a unique continuous map $\bar{f}$ with $f=\bar{f}\circ q$. The fibre calculation proves injectivity, and surjectivity follows from that of $f$. Hence $\bar{f}$ is the required continuous bijection.

The first integral singular homology of the Hawaiian earring has a direct summand isomorphic to $\mathbb{Q}$. This consequence of its structural homology theorem suffices, through the universal coefficient theorem for cohomology and nonzero Ext of the rationals with integer coefficients, to show that its integral $H^2$ is nonzero. The precise structural theorem is Theorem 3.1 of www.sas.rochester.edu/mth/sites/doug-ravenel/otherpapers/eda-kawamura2.pdf.