A homotopic equivalence of Torus Question:
Show that $\mathbb{R}^3 \setminus \{(0,0,z):z \in \mathbb{R}\} \ \bigcup \ \{(x,y,0):x^2+y^2=1\}$ is homotopic equivalent to Torus.
My thought:
I am trying to think of a geometric solution first.
Using the void in $z$-axis we can shrink the (almost) solid infinite cylinder $\{(x,y,z): x^2+y^2 \le 1\ , z \in \mathbb{R} , (x,y) \ne 0 , z\ne 0\} \setminus \{(x,y,0):x^2+y^2=1\}$ to the hollow cylinder $\{(x,y,z):x^2+y^2=1, z\ne 0\}$
Similarly we can also shrink the outside of the cylinder on its surface.
I cannot think any further. Please help me with this. If you try not to use heavy tools then it will be very helpful. As I am just starting to read Algebraic Topology.
 A: There's a lovely diagram of this homotopy equivalence at the math3ma blog here (it's #4). In the interest of not making this answer entirely trivial, I'll provide another way of seeing the equivalence here (though if I'm being entirely honest, I like her explanation more).
First, it should be clear that most of the "height" of $\mathbb{R}^3$ doesn't matter. Since we're cutting out the $z$ axis and stuff in the $xy$-plane, we can (and should) squish the $z$ axis down. Similarly, stuff that's "far away from the origin" doesn't matter. Since the $z$ axis and the unit circle are both close to the origin, we can contract everything outside of that region. That is, we can bring ourselves to this position:

Next, let's puff out the $z$ axis. We removed it, so we can make the hole bigger as long as we stay inside the other hole from the unit circle. Keeping with the rectangular theme, let's puff it out into a rectangular hole:

And now, of course, you see how we get a torus from this. All that's left is to remove the circle which is looped around the hole we just cut out. Of course, this circle lies entirely inside what's left of the box. So if we puff out the circle, we're carving out the inside of the box -- that is, we're hollowing out the donut. This leaves us with just the surface, which is a torus (albeit a rather boxy one).

I hope this helps ^_^
A: Consider a symmetric(w.r.t. ${XY}, {YZ},{ZX}$-planes) torus $T$ in $\Bbb R^3$ whose center circle is $S:x^2+y^2=1$. Consider two planes $P_1, P_{-1}$ in $\Bbb R^3$ which are parallel to $XY$-plane and touch $T$ in two circles, i.e. the torus $T$ is in between the planes $P_1, P_{-1}$. Now, $P_1,P_{-1}$ splits $\Bbb R^3$ into three portions, say $H_1:z\geq \varepsilon$ and $H_{-1}:z\leq -\varepsilon$ and $H_0:-\varepsilon \leq z\leq \varepsilon$ for some $\varepsilon >0$.

Define $A:=\Bbb R^3\backslash(Z\text{-axis}\cup S)$. Note that $H_{\pm 1}\backslash Z\text{-axis}$ has a strong deformation retract onto the punctured planes $P_{\pm 1}\backslash(0,0,\pm \varepsilon)$, respectively, considering vertical projections along $Z$-axis. So, $A$ has a strong deformation retract(see the definition below) onto $B:=H_0\backslash \big(L_\varepsilon\cup S\big)$, where $L_\varepsilon:=\{(0,0,z)\in \Bbb  R^3:|z|\leq \varepsilon \}$.

Now, we show there is a strong deformation retract of $B$ onto $T$.
$\bullet$ Let $p\in B$ be a point of the exterior part of $T$(note that $p$ may lie on $T$). The plane $\pi_p$ passing through $p$ and parallel to $XY$-plane intersects $T$ at most two circles. Let $\pi_p$ intersects $L_\varepsilon$ at the point $q$. Taking a radial projection on this plane $\pi_p$ w.r.t. the point $q$ one can project $p$ on the intersecting circle(s) defined by $T\cap \pi_p$. So, we have a deformation retract of $\text{exterior}(T)\backslash L_\varepsilon$ onto $T$. Note that this deformation retract is relative to $T$, i.e., each point on $T$ will be in its position in the whole process.
$\bullet$  Next, let $p\in B$ be a point of the interior part of $T$(note that $p$ may lie on $T$). Each plane in $\Bbb R^3$ of the form $\pi:ax+by+0\cdot z=0$ intersects the $\text{interior}(T)\backslash S$ in a punctured disk $D^*$ centered at some point $q\in S$. Considering radial projection on $D^*$ w.r.t. the point $q$ there is a deformation retract of $D^*$ onto its boundary $\partial D^*$. So, we have a deformation retract of $\text{interior }(T)\backslash S$ onto $T$. Again note that this deformation retract is relative to $T$, i.e., each point on $T$ will be in its position in the whole process.
Combining these, we have a strong deformation retract of $B$ onto $T$.

Finally, strong deformation retract a transitive notion, i.e., $A$ is strong deformation onto $B$, and $B$ is strong deformation onto $T$ imply $A$ is strong deformation retract onto $T$.

$\textbf{Definition:}$ A continuos map $H\colon A\times [0,1]\to A$ is said to be a strong deformation retract onto a subspace $B$, if $H(a,0)=a$ for all $a\in A$, $H(a,1)\in B$ for all $a\in A$ and $H(b,t)=b$ for all $(b,t)\in B\times [0,1]$.
