= Solution
Define <compactly supported cohomology> by
$$
\boxed{H_{ct}^q(X;A)=\varinjlim_{K\subseteq X\text{ compact}}H^q(X,X\setminus K;A).}
$$
For $K\subseteq L$, 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
$$
\boxed{H_{ct}^q(\mathbb R;\mathbb Z)=\begin{cases}\mathbb Z,&q=1,\\0,&q\ne1.\end{cases}}
$$
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. Thus
$$
H^q(X,X\setminus K)\cong H^q(Y,Y\setminus K).
$$
Complements 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 identifies
$$
H^q(Y,U)\cong\widetilde H^q(Y).
$$
In 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
$$
\boxed{H_{ct}^*(X)\cong\widetilde H^*(X^+).}
$$
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 gives
$$
\boxed{H_{ct}^q(\mathbb Z\times\mathbb R;\mathbb Z)=\begin{cases}\displaystyle\bigoplus_{j\in\mathbb Z}\mathbb Z,&q=1,\\0,&q\ne1.\end{cases}}
$$
The <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 \b[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> injects
$$
\operatorname{Ext}_{\mathbb Z}^1(H_1(W;\mathbb Z),\mathbb Z)\hookrightarrow H^2(W;\mathbb Z).
$$
The 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 $a_n-(n+1)a_{n+1}=0$, $n\geq1$. The corresponding free resolution shows that $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Q,\mathbb Z)$ is the cokernel of
$$
\prod_{n\geq1}\mathbb Z\longrightarrow\prod_{n\geq1}\mathbb Z,\qquad (u_n)\longmapsto(u_n-(n+1)u_{n+1}).
$$
The constant sequence $(1,1,\ldots)$ is not in the image. Otherwise iteration would give
$$
u_1=\sum_{j=1}^N j!+(N+1)!u_{N+1}\quad\text{for every }N.
$$
For 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 that
$$
\boxed{H_{ct}^2(\mathbb Z\times\mathbb R;\mathbb Z)=0,\qquad\widetilde H^2(W;\mathbb Z)\ne0,}
$$
which 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.
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