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.