Pictures of One-point compactification Are these the pictures of the following one-point compactification of the following surfaces:


*

*Two dimensional sphere with three points removed

*The disjoint union of two copies of $S^{1} \times \mathbb{R}$

*Disjoint union of $\mathbb{R}^{2}$ and $\mathbb{R}^{1}$

*Disjoint union of the real line together with an open cylinder, 


I am not sure about the second one (Should it be two cones joined at a point called $\infty$)and the 4th one shoud be a cone and a circle with the circle?

 A: Here is a result you could use:

Let $X$ and $Y$ be non-compact spaces and $Z=X\sqcup Y$ their topological sum. If $\hat V$ denotes the one-point-compactification of $V$ for $V=X,Y,Z$, then $\hat Z$ is homeomorphic to the wedge $\hat X\vee\hat Y$ at the respective points at infinity.

Proof: There is a map $f_X:\hat X\to\hat Z$ which sends $∞_X$ to $∞_Z$. On $X$ it is the open embedding $X\hookrightarrow Z↪\hat Z$, so $f$ is continuous on each point in $X$. For continuity on $∞_X$, let $V$ be an open neighborhood of $∞_Z$, that means its complement $Z-V$ is closed and compact in $Z$, thus $X-V$ and $Y-V$ are each compact and closed in $X,Y$ respectively. Therefore, $\hat X-(X-V)$ is an open neighborhood of $∞_X$ such that $f_X[\hat X-(X-V)]=V-Y\subset V$. So $f_X$ is continuous.
On the other hand, a closed and compact subset $K$ of $X$ is closed and compact in $X\sqcup Y$. That means that the image of an open neighborhood $\hat X-K$ of $∞_X$ under $f_X$ equals $(\hat Z-K)\cap (X\cup\{∞_Z\})$, a neighborhood of $∞_Z$ in $f\left[\hat X\right]$. Thus $f_X$ is a homeomorphism onto a closed image, i.e. a closed embedding.
Similarly, $f_Y$ is a closed embedding, so their disjoint union $f:\hat X\sqcup\hat Y\to \hat Z$ is a closed map. It thus induces a homeomorphism $\hat X\vee\hat Y\approx\hat Z$.
