# Clever Homotopy Equivalences

You know the routine. You come across a topological space $X$ and you need to find its fundamental group. Unfortunately, $X$ is an unfamiliar space and it's too difficult to look at explicit loops and relations. So what do you do? You look for another space $Y$ that is homotopy equivalent to $X$ and whose fundamental group $\pi_1(Y)$ is much easier to compute. And voila! Since $X$ and $Y$ are homotopy equivalent, you know $\pi_1(X)$ is isomorphic to $\pi_1(Y)$. Mission accomplished.

Below is a list of some homotopy equivalences which I think are pretty clever and useful to keep in your back pocket for, say, a qualifying exam or some other pressing topological occasion.

## #1: A sphere with n points identified (n > 1)

This space is homotopic to the wedge sum of a sphere and $n-1$ circles. It helps to start by looking at the case when $n=2$, and there's no harm in letting those two points be the 'north and south pole' of the sphere. To identify the points, we could just smush them together, but we'd only end up with a chubby bagel and as cute as that may be, it won't help us find the fundamental group (our ultimate goal). Instead, we can identfy the two points as shown on the right below:
But what is its fundamental group? To answer that, we use the general fact that for any CW-complex $X$ with a contractible subcomplex $A$, the quotient space $X/A$ is homotopic to $X$. (See page 11 here.) In particular, let $X$ be the space shown below, where $B$ is an arc along the sphere joining the north and south poles, and $A$ is also an arc which joins them but which lies off of the sphere. Note that $X$ does indeed have a CW-complex structure, with the endpoints of $A$ and $B$ being 0-cells, $A$ and $B$ themselves are 1-cells, and the surface of the sphere is a 2-cell.

Contracting $A$ to a point gives us $X/A$ which is precisely our original space above. But we can also contract $B$ to a point to obtain $X/B$ - the wedge sum of a sphere and circle! Thus the homotopy equivalence $X/A\simeq X \simeq X/B$ shows that our original space is homotopic to $S^1\vee S^2$ whose fundamental group can easily be shown to be $\mathbb{Z}$.

The case for $n>2$ is simiply a generalization of the above. For illustration let's suppose $n=4$.

Here we can imagine $A_1,A_2$ and $A_3$ as passing through the inside of the sphere (it's just easier to draw this way) and $B_1,B_2$ and $B_3$ as lying on the surface. If we contract each of the $B_i$'s, we end up with the wedge of a sphere and three circles. But we could also have contracted each of the $A_i$'s instead. This would give us a sphere with four points identified (which is harder to draw!):

Quoting the theorem we used in the case when $n=2$, we see that a sphere with four points identified is homotopy equivalent to $S^2\vee S^1\vee S^1\vee S^1$. From here it's not too hard to see that for $n>1$, $$S^2/\{n \text{ points}\} \simeq S^2\vee \overbrace{S^1\vee\cdots \vee S^1}^{n-1}$$

## #2: A torus with n points identified (n > 1)

Given our approach to #1, you might guess that identifying $n$ points on a torus gives a space homotopic to the wedge sum of a torus with $n-1$ circles* - and you'd be right! In fact, the intuition is nearly identical to what we did in #1, so I shan't recap it here.

## #3: ℝ³\{circle}

This space is homotopic to $S^2\vee S^1$. To see this, start by drawing $\mathbb{R}^3$ as a solid cube (just to help us visualize things) and imagine removing a circle from it. You can see this in the drawing below where $X=\mathbb{R}^3\smallsetminus\{\text{circle}\}$ consists of the space inside the yellow cube including the blue disc, but not including the white boundary of that disc. (That white part is our deleted circle which I will from now on refer to as "our/the circle.") Notice that our circle sits on some plane in $\mathbb{R}^3$. (I've shaded it in dark yellow.) Then all the points lying on that plane outside of the circle (the dark yellow bit) along with the points in $\mathbb{R}^3$ not on the plane (the light yellow bit) deformation retract onto a sphere. (So basically, everything that's yellow maps to a sphere.) And all the points enclosed by the circle (the blue disc) deformation retract onto a line, i.e. the diameter of our sphere. Then we can slide one end of that diameter along the sphere to obtain the wedge of a circle with a sphere!
With this in mind, it's not too difficult to see that the complement of two (unlinked) circles in $\mathbb{R}^3$ is homotopic to $(S^2\vee S^1)\vee(S^2\vee S^1)$.

Added May 22, 2016: I realize (Thank you, readers!) that the description of the first homotopy equivalence in the picture above is a bit cryptic. And to make matters worse, that horizontal blue line should really be vertical! Yikes. So instead of trying to illuminate (and thus possibly obscure) the situation with more words, let me show you a drawing of what's really going on (at least, in my brain). You'll have to excuse my shabby artistic skills, but hopefully the deformation retract is a little clearer now:

## #4: ℝ³\{z-axis ∪ circle centered at origin}

Thinking again of $\mathbb{R}^3$ as a cube, this space - let's call it $X$ - is the complement of the white line and white circle below:

Believe it or not, this is homotopy equivalent to a torus! To see this, first recall that $\mathbb{R}^2\smallsetminus\{\text{a point}\}$ is homotopy equivalent (i.e. it deformation retracts) to a circle.

Now within our space $X$, imagine lining up a plane along the $z$-axis which is perpendicular to the $xy$-plane. This plane, drawn in yellow below, is missing a point where the (deleted) circle centered at the origin "punctures" it. This means that our plane can deformation retract to a circle as shown below:

Of course, there are infinitely many planes such as the one above! So repeating this plane-to-circle process these infinitely many times, we obtain the desired torus.

Voila!

## #5: A torus with n points removed

This one is actually pretty easy - it's homotopic to a wedge of $n+1$ circles! It helps to begin by drawing the torus as a CW-complex, i.e. a rectangle with opposite edges identified. In the case when $n=1$, our space looks like a rectangle with a disc removed. That object can then deformation retract onto its "wire frame" which, after identifying the edges as labeled, gives us a wedge of two circles:

But it's the same exact same thing for $n>1$! Here's what it looks like when $n=4$:

## #6: ℝ³\{two linked circles}

Believe it or not, this space is homotopic to the wedge sum of a sphere and a torus! This deformation retraction is a little more difficult to describe than the ones above, so I'll just give the hand-wavy version. Again it's useful to imagine $\mathbb{R}^3$ as a cube, and let us presciently place the $S^2\vee T^2$ inside the cube to make things easier for us. Notice that there are three regions we need to worry about: #1 outside the sphere, #2 inside the torus, and #3 inside the sphere but outside the torus. The first region is easy enough to deal with - simply send all points outside the sphere to the sphere's surface. This is the deformation retraction shown to the right, below:
Now recall that our sphere is a solid sphere minus the two linked circles, so region #2 - the torus - is a solid torus, minus one of those circles. So the next step is to simply deformation retract all points inside region #2 outwards to the surface of the torus. At this point it's helpful to pretend the surfaces of both the torus and the sphere are made of a rigid material like hard plastic. So now, when we deal with region #3, imagine the complement of the torus inside the sphere to be made of pliable material like soft clay.
Finally, note that region #3 contains a circular-shaped hole, namely the second of the omitted circles. (Referring to my drawing, this is the rightmost one. It's the more elliptical/ovally looking of the two.) We can slowly stretch/enlarge that hole until it can't get any bigger, i.e. until it contacts the hard surface of the sphere and the torus. Essensially, this is just 'hollowing-out' the the part of the sphere that's outside the torus. And that's it! The remaining space is $S^2\vee T^2$.

*Edited May 11, 2016: This originally (and incorrectly!) read "n-1 spheres." My sincerest thanks to a reader for pointing this out!