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I posted a similar but different question before in the link

https://math.stackexchange.com/questions/4311982/why-does-x-0-times-s1-simeq-x-x-0/4312530?noredirect=1#comment8987557_4312530.

Now, my new question is the following:

enter image description here

My question is as follows:

In the above picture, it says that $X_0\times S^1\rightarrow X-X_0$ is a homotopy equivalence which is obtained by taking the boundary of the small tubular neighborhood of $X_0$. I'm confused about how to understand the inclusion just by taking the boundary of the small tubular neighborhood of $X_0$. What is that? And Can we guarantee the existence of homotopy equivalence? If so, take an example.

I'm not sure if this question is too easy for math overflow. I put it on stack exchange last week but nobody answered it. Could you please help me? Thanks in advance.

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    $\begingroup$ Since $X_0$ has codimension 2, a tubular neighbourhood looks like $X_0\times B$, where $B$ is the 2-dimensional disk. Then the bounday of that tubular neighbourhood is $X_0\times \partial B$ and $\partial B$ is the circle. $\endgroup$
    – user130903
    Commented Nov 23, 2021 at 8:43
  • $\begingroup$ Do you have examples of the condition of homotopy equivalence to let the theorem make sense? $\endgroup$
    – Radeha Longa
    Commented Nov 23, 2021 at 8:49
  • $\begingroup$ Yes, the embedding of ${\mathbb R}^{n-2}$ into ${\mathbb R}^n$. $\endgroup$
    – user130903
    Commented Nov 23, 2021 at 12:42

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