Is this Mobius strip definition correct? $I = [0, 1] \subseteq \mathbb{R}$ $I^2 = I \times I \subseteq \mathbb{R^2}$ with the usual topology Then the Mobius strip, $M$ is $I^2 /\sim$ where $\sim$ is the following equivalence given by $x \sim y \iff$ $x = y$ or $x = (0, t), \ y = (1, 1-t)$ for some $t \in I$ or $x = (1, 1-t), \ y = (1, t)$ for some $t \in I$ Is that definition correct?

No, that isn't correct. You want $x\sim y$ if $x=y$ or $x=(0,t)$ and $y=(1,1-t)$ or $x=(1,t)$ and $y=(0,1-t)$.

You start with a rectangle, and "glue" two sides together with a twist.

The equivalence $(1,t)\sim(1,1-t)$ is a bit strange. Indeed, what you have defined is not an equivalence relation, since $(0,0)\sim(1,1)\sim(1,0)$ in your definition, but it is not true that $(0,0)\sim(1,0)$. But even if you extend your relation to an equivalence relation, what you get is something that is a Mobius strip with one cross-line folded. In particular, it is not a "manifold with boundary," since no neighborhood of the equivalence class of $(0,1/4)$ is homeomorphic to a disk.