Torus without a disk I have troubles to visualize the torus $\mathbb{T}^2$ without a disk $\operatorname{Int}D^2$. I don't see why $\mathbb{T}^2\setminus \operatorname{Int}D^2$ is not the same as $\mathbb{T}^2\#\mathbb{T}^2$?
If I form figure $2$ (the right) I should be able to get figure $1$ (the left). So I don't really see the difference?
Also in my solution it is mentioned that $\mathbb{T}^2\setminus \operatorname{Int}D^2$ is the same as square where one edge is removed. A CW-Complex with one $0-$cell, three $1-$cell and one $2-$cell. Unfortunately I don't see this.
Unfortunately I don't really find any helpful pictures or literature where $\mathbb{T}^2\setminus \operatorname{Int}D^2$ is mentioned.
So if anyone could explain this to me, I would be thankful.

 A: Let $I=[0,1]$ and let $\sim$ be the equivalence relation generated by: $$\forall x\in I,(x,0)\sim (x,1),\forall y\in I,(0,y)\sim (1,y).$$
Then, $\mathbb{T}^2$ is homeomorphic to $I^2/\sim$. Let $\pi: I^2\twoheadrightarrow\mathbb{T}^2$ be the canonical surjection.
Removing a disk or a point from $\mathbb{T}^2$ leads to homeomorphic results. Let us see that $\mathbb{T}^2\setminus\{\pi(0)\}$ is a deformation retract of $\mathbb{S}^1\wedge\mathbb{S}^1$. First, notice that $I^2\setminus\{0\}$ is a deformation retract of $\partial I^2$. Indeed, let define the map $f\colon I^2\setminus\{0\}\rightarrow\partial I^2$ by for any $x\in I^2\setminus\{0\}$, $f(x)$ is the intersection of $[0,x)$ with $\partial I^2$. Then, $f$ is continuous and $f_{\vert\partial I^2}=\textrm{id}_{\partial I^2}$. Therefore, $(f_s)_{s\in[0,1]}$ define by : $$\forall s\in[0,1],f_s:x\mapsto (1-s)x+sf(x),$$
is a deformation retraction of $I^2\setminus\{0\}$ to $\partial I^2$. As a last step, notice that for any $s\in[0,1]$, $\pi\circ f_s$ is constant on the equivalence classes of $\sim$ and leads to $(\overline{f}_s)_{s\in [0,1]}$ a deformation retraction of $\mathbb{T}^2\setminus\{\pi(0)\}$ to $\partial I^2/\sim$. To conclude, notice that $\partial I^2/\sim$ is a wedge sum of two circles.
$\mathbb{T}^2$ minus a disk is a deformation retract of a wedge sum of two circles. In particular, it cannot be homeomorphic to $\mathbb{T}^2\#\mathbb{T}^2$, since they don't have the same fundamental group.
Remark. There is still some work to do, but the main idea is that $I^2$ minus a point is a deformation retract of $\partial I^2$ and that $\partial I^2/\sim$ is a wedge sum of two circles. The rest of it is only formalism, in particular the use of the universal property of the quotient topology.
