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What is the one point compactification of $S^n\times \mathbb{R}$?

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    I have a naive guess that it may be homotopy equivalent to $S^{n+1}\vee S^1$.2011-04-06
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    Do you have a guess? Can you think of a compact space which contains a point such that its complement is homeomorphic to $S^n\times\mathbb R$?2011-04-06
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    @user8484: are you looking for the definition of the one-point compactification? That's how I read your question. But your comment suggests you're looking for a more elementary description of the homotopy-type.2011-04-06
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    For example, if you had asked "what is the real numbers?" tends not to indicate the questioner is interested in a homotopy-type description.2011-04-06
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    $S^n\times S^1/S^n\times\{\text{pt}\}$2011-04-06

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It might be helpful to think of this as $\mathbb{S}^n \times [0,1]$ with the boundary identified to a point. It's easy to see that $\mathbb{S}^n \times \mathbb{R}$ embeds homeomorphically as an open dense set by the product of the identity on $\mathbb{S}^n$ with any homeomorphism from $\mathbb{R}$ to $(0,1)$. Since the complement of the image of $\mathbb{S}^n \times \mathbb{R}$ is a singleton, this is a one-point compactification of $\mathbb{S}^n \times \mathbb{R}$.

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    Notice that, if you only collapse $\mathbb S^n\times\{0\}$ to a point $p$ and collapse $\mathbb S^n\times\{1\}$ to another point $q$, the result is homeomorphic to $\mathbb S^{n+1}$. Now finish the collapsing described in this answer, by identifying $p$ with $q$. So the final space can be described as $\mathbb S^{n+1}$ with two points collapsed to one.2016-09-15
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    @AndreasBlass That's interesting, the first statement is not intuitive to me. How would you prove it?2016-09-15
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    Wait, I think I can see it now. The "top half" corresponds to the upper hemisphere, while the "bottom half" corresponds to the lower hemisphere. Makes sense.2016-09-16