# Re-defining a map after annulus removal from the torus

Let $$\Bbb T$$ be the torus $$\Bbb S^1\times \Bbb S^1$$ and $$f\colon \Bbb T\to \Bbb T$$ be a map. Suppose we have embeddings $$j\colon \Bbb S^1\hookrightarrow \Bbb T$$ and $$\varphi\colon \Bbb S^1\times [0,1]\hookrightarrow \Bbb T$$ such that $$f\circ \varphi(e^{2\pi i\theta},r)=j(e^{2\pi i\theta})$$ for all $$r\in [0,1]$$ and for all $$\theta\in [0,2\pi]$$. Let $$A\subseteq \Bbb T$$ be the annulus given by $$\varphi$$, i.e. $$A=\text{im}(\varphi)$$.

Can we re-define $$f$$ as a map from $$\Bbb T\to \Bbb T$$ after removing the interior of the annulus $$A$$ and the pasting the two boundary components $$A$$?

This seems visually obvious to me as using the hypothesis on $$f$$ one can re-define $$f$$ from $$\Sigma:=\frac{\displaystyle \Bbb T\backslash \varphi\big(\Bbb S^1\times (0,1)\big)}{\displaystyle\varphi(e^{2\pi i\theta},0)\sim \varphi(e^{2\pi i\theta},1)}$$ into $$\Bbb T$$. What is not clear to is that whether $$\Sigma$$ is orientable or not? We have two possibilities: either $$\Sigma$$ is homeomorphic to Torus or Klein bottle.

The problem comes when I consider the following fact:

Let $$M$$ be a connected oriented manifold of dimension $$n$$ and $$\partial M=\partial_+M\ \sqcup \partial_-M$$ such that there is an orientation preserving homeomorphism $$g\colon\partial_+M\to \partial_-M$$(consider the induced orientation on $$\partial M$$). Then the manifold $$N:=\frac{\displaystyle M}{\displaystyle a\sim g(a)}$$ is not orientable.

• The two boundary components of $S^1\times[0,1]$ have opposite orientation (effectively, since the endpoints of $[0,1]$ have opposite orientation). Sep 20 '21 at 8:49
• Oh, nice! Are you saying this thing if I give standard orientation on $\{1\leq |z|\leq 2\}$ comes from $\Bbb C$ then the orientations of one of the circles $|z|=1, |z|=2$ will be clockwise and other will be anti-clockwise. Sep 20 '21 at 8:54
• Yes, e.g. if your convention for boundary orientation is an "outward-pointing last" convention, the outer circle will have the same boundary orientation as it does usually, i.e. when you consider it as boundary of a disk. But the outward-pointing tangent vectors on the inner circle point inwards on the disk $\{|z|\le1\}$, which has $\{|z|=1\}$ as boundary, so that one is oriented the opposite way. Sep 20 '21 at 9:04
• So, the identification given in the numerator of the definition of $\Sigma$ is actually by an orientation reversing map $g\colon \partial_+A\to \partial_-A$, in other words, $\Sigma$ is orientable i.e. $\Sigma\cong \Bbb T$. Sep 20 '21 at 9:11
• Yes. However, it's worth pointing out this conclusion can also be reached completely explicitly: the embedding $\varphi$ extends to an embedding of $S^1\times(-\varepsilon,1+\varepsilon)$ for some small $\varepsilon>0$ by choosing a small tubular neighborhood. The result of removing $S^1\times[0,1]$ from $S^1\times(-\varepsilon,1+\varepsilon)$ and then gluing the boundary components is diffeomorphic to $S^1\times(-\varepsilon,1+\varepsilon)$ again (draw a picture, but you can also give a formula explicitly) and this diffeomorphism can be chosen to have compact support. Sep 20 '21 at 10:19