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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.

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    $\begingroup$ The two boundary components of $S^1\times[0,1]$ have opposite orientation (effectively, since the endpoints of $[0,1]$ have opposite orientation). $\endgroup$
    – Thorgott
    Sep 20 '21 at 8:49
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    $\begingroup$ 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. $\endgroup$
    – Random
    Sep 20 '21 at 8:54
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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Sep 20 '21 at 9:04
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    $\begingroup$ 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. $\Sigma\cong \Bbb T$. $\endgroup$
    – Random
    Sep 20 '21 at 9:11
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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Sep 20 '21 at 10:19

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