Here's my motivation for this question: Given the sphere $S^2$, we can "puncture" it by removing a single point, obtaining $S^2 \setminus \{P\}$, which is homeomorphic to $\mathbb{R}^2$. Conversely, given $\mathbb{R}^2$, we can construct a space homeomorphic to $S^2 \setminus \{P\}$ through a projection map into $\mathbb{R}^3$. We can establish similar homeomorphisms between $S^n \setminus \{P\}$ and $\mathbb{R}^n$.
The process of "puncturing" can be very easily generalized: given a topological space $X$, one can puncture it by creating the subspace $X \setminus \{P\}$ for any point $P \in X$. We end up with a new topological space with interesting properties and connections to the original, and if it doesn't matter which point is removed, we can refer to this space as "the punctured $X$." But the process of "unpuncturing" — going from $\mathbb{R}^n$ to $S^n$, for example — is not nearly as obvious. Naively adjoining a single point to form the space $X \sqcup \{P\}$ doesn't work, because $P$ does not necessarily share the same local properties as the other points in this space.
Put formally, my question is this:
Let $X$ be a homogeneous topological space (that is, for every two distinct points $P, Q \in X$, there exists a self-homeomorphism $X \to X$ sending $P$ to $Q$). Does there exist a homogeneous topological space $Y$ such that the punctured $Y$ is homeomorphic to $X$? If so, is this space unique, and how can one construct it? If not, what is a valid counterexample?
(The self-homeomorphism property is my way of making sure that it doesn't matter which point is removed. This property is enjoyed by all of the nice topological spaces that come to mind, like path connected manifolds without boundary.)
On one hand, this property doesn't feel particularly restrictive: constructing such a space $Y$ would only require finding an appropriate way to define the open neighborhoods of $P$ in the space $X \sqcup \{P\}$. But on the other hand, I can't think of any obvious way to "unpuncture" most topological spaces, like the $n$-sphere or any compact manifold.
(Note: my topology background is fairly limited. I know enough point-set topology to speak the language, and I know a thing or two about the basic algebraic structures on topological spaces, but I certainly have no level of specialization in the subject. I would be extremely surprised if this question hasn't already been asked and thoroughly answered, if perhaps in a different manner, but I don't really know where to find the relevant theory.)
Best Answer
Thanks to Ulli's comment, I have worked out an answer to the formal question, which for posterity's sake I'll write in detail here.
The answer is no: there is no such $Y$ for every homogeneous space $X$. In particular, if a homogeneous space $X$ is also compact, Hausdorff, and infinite (in cardinality), then there is no such $Y$.
Suppose such a $Y$ existed. We first claim that $Y$ is also Hausdorff. Indeed, let $x,y \in Y$. Since $Y$ is infinite, we may choose a $P \in Y$, $P \ne x,y$. Because $Y \setminus \{P\} \cong X$, we may identify $X$ with $Y \setminus \{P\}$, and then use the fact that $X$ is Hausdorff to create two open neighborhoods $U_x$, $U_y$ of $x$, $y$ respectively, such that $U_x$ and $U_y$ are disjoint on $Y \setminus \{P\}$. If $P \in U_x \cap U_y$, then $U_x \setminus \{P\}$ is still open because $\{P\}$ is closed (by any homeomorphism $X \cong Y \setminus \{Q\}$, $Q \ne P$, and the fact that $X$ is Hausdorff $\implies$ every point-set in $X$ is closed). Thus we can find disjoint open neighborhoods separating $x$ and $y$.
We now show that any point $P \in Y$ is open as a point-set. Fix this $P$ and let $x$ vary over $Y \setminus \{P\}$. Since $Y$ is Hausdorff, there exist disjoint neighborhoods $U_x$, $V_x$ of $x$, $P$ respectively. Then $\{U_x\}$ forms an open cover of $Y \setminus \{P\}$, and so admits a finite subcover since $X \cong Y \setminus\{P\}$ is compact. Denote this open subcover as $U_1, U_2, \dots, U_n$, and for each $U_i$ let $V_i$ denote the corresponding disjoint open neighborhood of $P$. Then $\bigcup_{i=1}^n U_i \supseteq Y\setminus \{P\}$, i.e. $\bigcap_{i=1}^n U_i^c \subseteq \{P\}$. And since $U_i \cap V_i = \varnothing$ for all $i$, $V_i \subseteq U_i^c$, and therefore $$\{P\} \subseteq \bigcap_{i=1}^n V_i \subseteq \bigcap_{i=1}^n U_i^c \subseteq \{P\}.$$
Thus $\{P\} = \bigcap_{i=1}^n V_n$ which is a finite intersection of open sets and is therefore open.
$\{P\}$ is thus both open and closed for every $P \in Y$. But then $Y$ is discrete, and so $X$ is discrete as well. This contradicts the compactness of $X$, since the open cover $\{\{x\}\}_{x \in X}$ would admit no finite subcover.