Deforming the torus without a point to $S^1 \lor S^1$ Let $T$ be the topological torus, given by taking a rectangle with parallel sides oriented in the same direction, and glueing together each pair of parallel sides along the given direction.

Take a point $P\in T$ and remove it, I want to show that $S^1 \lor S^1$ is
a deformation retract of $T-\{P\}$ which is given by glueing two
circles at some point.

Intuitively:

*

*The point $P$ divides the inner area of the rectangle in four parts, those points lying above $P$ or below $P$, and left and right of $P$.

*These parts will be the triangles delimited by the segments connecting the point $P$ to the vertices.

*Now I can define a retraction collapsing each point of the rectangle to the corresponding point on the base of the triangle it belongs to.
This is not well defined for points lying on the segments connecting $P$ to the vertices, but each choice will give some retraction, and the retraction will be homotopic to the identity because the rectangle is a convex set (edit: I think this point is wrong, because if I remove a point I lose the convexity).

*So we have produced a retraction of the rectangle onto its sides.

*Passing to the quotient which identifies the parallel sides, this composition is still a retraction if we quotient also the rectangle by the same relation making it a torus.

*We conclude by observing that quotienting the boundary of the rectangle by the above relation gives a space homeomorphic to two circles glued at a point, and we are done.


Can we make this argument more rigorous?
What I am not satisfied about my argument is that it does not make
clear why it is important to remove a point from the torus. What is a
rigorous way to make clear that removing a point is necessary to make
this argument work?

My main question is:

Where precisely in the argument above am I using the fact that I removed the point $P$?

 A: I think everything has been resolved in the comments already, but here is a bit more information. The main point is that a two-dimensional square with its centre removed can be deformation retracted onto its perimeter, see for instance the linked post by Angina Seng. The details of how this is accomplished aren't too important. For instance, Tyrone outlined in the comments how to do this by first identifying the square with the disk. Just for the purposes of providing another perspective, here is an approach you could use to get things down to the level of formulas without passing from the square to the Euclidean disk.
Let $(X,\| \cdot\|)$ be a normed vector space. Let $B = \{ x \in X : \|x\| \leq 1\}$ be the closed unit ball and $S= \{x \in X: \|x\|=1\}$ the unit sphere. Its quite straightforward to give a homotopy $f_t:B \setminus \{0\} \to B \setminus \{0\}$ deforming the identity map of the punctured ball into the "normalization map" $x \mapsto \frac{x}{\|x\|} : B \setminus \{0\} \to S$ while keeping the points of $S$ fixed. Just divide a vector $x$ by a factor which continuously changes from $1$ to $\|x\|$, say
$$f_t(x) = \frac{x}{1-t + t\|x\|}.$$
It is important to see that the normalization map does not extend to the whole ball, since one gets a division by zero in that case. Removing a point is crucial to getting the retraction.
Now, taking $X= \mathbb{R}^2$ and using the $\infty$-norm
$$\|x\|_\infty = \max(|x_1|,|x_2|)$$
one has that the ball and sphere are, respectively, the 2-dimensional square of side length 2 centred at $0$ and the perimeter of that square, so above procedure applies to this case, in particular.
A: Let me answer your main question by first considering a different situation.
Deforming the sphere $S^2$ with a point to a point: Take a point $P \in S^2$ and remove it, and show that what's left deformation retracts to a point. To do this let's choose $P$ to be the north pole $P=(0,0,1)$, and we'll define a deformation retraction of $S^2 - \{P\}$ to the south pole $Q = (0,0,-1)$.
Intuitively, the deformation restriction moves each point of $S^2 - \{P\}$ southward, along the longitude line through that point, to the south pole $Q$.
Where precisely in this argument did we use the fact that we removed the point $P$?
There is no well-defined longitude line through the north pole; in some sense the north pole lies on every longitude line. Thus, we had to remove the north pole before our deformation retraction could be well-defined.
What is a rigorous way of doing this?
Use spherical coordinates in $\mathbb R^3$, whose required properties, including apprpriate continuity properties, are known to you from your knowledge and expertise of analytic geometry. Using spherical coordinates, write down a formula for the deformation retraction
$$h : (S^2 - \{P\}) \times [0,1] \to S^2 - \{P\}
$$
The formula for $h$ that you write down should have the effect that the latitude coordinate (usually in $[0,2\pi]$, with $0$ and $2\pi$ identified) does not change as the time parameter $t \in [0,1]$ increases from $0$ to $1$. But the longitude coordinate (usually in $[-\pi/2,\pi/2]$ with $-\pi/2$ as the south pole and $+\pi/2$ as the north pole) should decrease at constant speed from its initial value in $[-\pi/2,\pi/2)$, moving along its latitude line to the final value $-\pi/2$.
Notice: the north pole had to be omitted because it does not lie on a well-defined longitude, and so there is no way to extend the formula for $h$ continuously. Intuitively, we cannot continuously choose a longitude line along which the north pole moves down toward the south pole. While it's also true that the longitude line at the south pole is not well-defined, the south pole does not move under the deformation retraction.
But,  for full rigor you must actually write down the formula for $h$, and check all of its the required properties for the desired deformation retraction.

Now, on to the torus.
Model the torus $T$ as the quotient of the square $R = [-1,+1] \times [-1,+1]$ with respect to the equivalence relation generated by $(x,-1) \sim (x,+1)$ and $(-1,y) \sim (+1,y)$. In place of cylindrical or spherical coordinates in $S^2$, use radial coordinates on the square $R$. Its boundary $\partial R$ is the union of the four sides $\{-1\} \times [-1,+1]$, $\{+1\} \times [-1,+1]$, $[-1,+1] \times \{-1\}$, $[-1,+1] \times \{+1\}$. We are going to remove the point $\mathcal O = (0,0)$. Using our knowledge and expertise in plane analytic geometry, each point $x \in R - \{\mathcal O\}$ can be written uniquely in the form
$$r(x) \cdot b(x)
$$
where
\begin{align*}
r(x) &= \frac{1}{\max\{x_1,x_2\}} \\
b(x) = \frac{x}{r(x)}
\end{align*}
We have removed the point $\mathcal O$ in order that these expressions $r(x)$ and $b(x)$ be well-defined and continuous as functions of $x \in R - \mathcal O$.
Now use the coordinates to define the formula for the deformation retraction
$$h : (R - \mathcal O) \times [0,1] \to R - \mathcal O
$$
Intuitively, the formula for $h$ keeps the boundary coordinate $b(x)$ constant, whereas the "radial" coordinate increases linearly from its initial value $r(x) \in (0,1]$ to its final value $1$, as $x$ moves outward along its radial segment towards $\partial R$.
Notice: the central point $\mathcal O$ had to be removed, because it does not lie on a well-defined radial segment and therefore there is no way to extend $h$ continuously. Now, under the identification of $\partial R$ to a wedge of two circles, a point in that wedge does not correspond to a well-defined point of $\partial R$, instead it correspond to either $2$ or $4$ points of $\partial R$; however, this does not matter because those points do not move under the deformation retraction.
