What happens if we remove the point of origin from the moebius band?

Question

What happens if we remove the point of origin from the moebius band ( using the deficion of the moebius band in $[0,1]\times[0,1]$)? what topological space do we get? I know that if we remove the origin of the torus (using the deficion of the moebius band in $[0,1]\times[0,1]$) then what I have left is homotopically equivalent with $\mathbb{S}^1\vee\mathbb{S}^1$, but in the moebius band what would it be? Thank you.

Answer 1

I'll presume that by the "definition of the mobius band in $[0,1] \times [0,1]$" you mean that one should identify $(0,t) \sim (1,1-t)$ for each $t \in [0,1]$.

I'll also presume that by the "point of origin" you mean the point $(0,0) \in [0,1] \times [0,1]$.

Under those presumptions, what you get is the moebius band with one point removed from its boundary circle. Like the moebius band itself --- which is homotopy equivalent to $\mathbb S^1$ --- the result of removing one point from the boundary circle is still homotopy equivalent to $\mathbb S^1$. In fact, removing a point from the boundary of any manifold-with-boundary leaves the homotopy type unchanged.

On the other hand, suppose that you had asked what you get when removing the point $\left(\frac{1}{2},\frac{1}{2}\right)$, which would be the same as removing a point from the interior of the moebius band itself. Then the result would be homotopy equivalent to $\mathbb S^1 \vee \mathbb S^1$.