Hatcher 3.3.21: What is a cone? I'm trying to solve exercise 3.3.21 from Hatcher's Algebraic Topology, and I'm a bit stuck on not quite understanding what he means by cone and cone point:

For a space $X$, let $X^{+}$ be the one-point compactification. If the added point, denoted $\infty$, has a neighbourhood in $X^{+}$ that is a cone with $\infty$ the cone point, show that the evident map $H^{n}_c (X;G) \rightarrow H^n (X^{+}, \infty; G)$ is an isomorphism for all $n$. [Question: Does this result hold when $X=\mathbb{Z} \times \mathbb{R}$?]

Okay, so, I would infer from the question that in the one-point compactification of $\mathbb{Z} \times \mathbb{R}$, $\infty$ has no such neighbourhood that is a cone. Thing is though, why not? My naive assumption was that by a cone shaped neighbourhood, Hatcher was simply meaning that there existed some space $Y$ such that if the neighbourhood in question was $N_{\infty}$, then $CY$ is homeomorphic to $N_{\infty}$ and if the homeomorphism is denoted $\phi$ and $p$ is the "tip of the cone", then $\phi(p) = \infty$.
But then it would appear to me that in the one-point compactification of $\mathbb{Z} \times \mathbb{R}$ there definitely exists a cone-shaped neighbourhood of $\infty$, specifically, one that is homeomorphic to the cone of $\mathbb{Z}$.
Am I doing something seriously wrong here? If so where? How am I to interpret what Hatcher has written?
 A: Your interpretation is correct, but there does not exist a cone-shaped neighborhood of $\infty$ in the 1-point compactification of $\mathbb{Z}\times\mathbb{R}$.  If you take a subset of the 1-point compactification of the form $A=((-\infty,-1]\cup[1,\infty))\times\mathbb{Z}\cup\{\infty\}$ then there is a continuous bijection $C\mathbb{Z}\to A$ from the cone on $\mathbb{Z}$ (sending the cone over each point of $\mathbb{Z}$ to one of the intervals heading out to $\infty$).  However, there are two issues that prevent this from being a cone-shaped neighborhood.
First, $A$ is not even a neighborhood of $\infty$ in the 1-point compactification.  In the 1-point compactification topology, any neighborhood of $\infty$ must contain a cocompact subset of $\mathbb{Z}\times\mathbb{R}$, which must contain the entirety of $\{n\}\times\mathbb{R}$ for all but finitely many $n\in\mathbb{Z}$.  So $A$ does not contain any neighborhood of $\infty$, and indeed every neighborhood of $\infty$ has to contain infinitely many full copies of $\mathbb{R}$ which turn into circles when the point at $\infty$ is added.  (The 1-point compactification can be identified with the notoriously pathological Hawaiian earring space.)
Second, even if you were taking a 1-point compactification of $\mathbb{Z}\times[0,\infty)$ instead of $\mathbb{Z}\times\mathbb{R}$ (so the entire 1-point compactification would really just look like a cone on $\mathbb{Z}$ without any circles), the continuous bijection from $C\mathbb{Z}$ would still not be a homeomorphism.  As above, every neighborhood of $\infty$ in the 1-point compactification would have to completely contain all but finitely many of the edges going out to $\infty$.  On the other hand, the cone topology is a quotient of the product topology on $\mathbb{Z}\times[0,1]$, so a neighborhood of the cone point just has to contain some open interval around the cone point on each of its edges (which can be chosen completely independently of each other).
