# 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)

*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:

*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.

*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 $n=4$.

## #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}

*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!

*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}

Voila!

## #5: A torus with n points removed

## #6: ℝ³\{two linked circles}

*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:

*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.

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