Showing that a space does not deformation retract

Question

What is a good way to show that a space does not deformation retract onto something? For example,

1. Show that the Mobius strip does not deformation retract onto its boundary
2. Show that the torus does not deformation retract onto the one-point union of two circles

Answer 1

One general way to prove nonexistence of maps is to assume the map exists, apply one of your favorite algebraic topology functors, and thereby obtain a contradiction. That's how the proof of the "no retraction" theorem goes for nonexistence of a retraction from a closed 2-dimensional disc to its boundary circle, a proof which you probably know (and which is a key step in proving the Brouwer fixed point theorem).

In this situation, suppose that $M$ is a Möbius band with boundary circle $\partial M$. To say that a function $f : M \mapsto \partial M$ is a retraction means that the composition $$\partial M \xrightarrow{i} M \xrightarrow{f} \partial M $$ is equal to the identity map, where $i$ is the inclusion. Now let's apply our favorite functor, the fundamental group functor. It follows that the composition $$\underbrace{\pi_1(\partial M)}_{\mathbb Z} \xrightarrow{i_*} \underbrace{\pi_1(M)}_{\mathbb Z} \xrightarrow{f_*} \underbrace{\pi_1(\partial M)}_{\mathbb Z} $$ is equal to the identity map. What could possibly go wrong?

Well, we know that $i_*$ is the "times $2$" homomorphism of $\mathbb Z$, because the fundamental group of $M$ is generated by the curve that goes around the core of $M$, and the boundary of $M$ is homotopic to the curve that goes $2$ times around the core.

Now if you work through the algebra, you will discover that no matter what homomorphism of $\mathbb Z$ you compose after the "times $2$" homomorphism, you will never get the identity homomorphism of $\mathbb Z$. That's the contradiction which proves that you cannot retract $M$ to $\partial M$.

As you study algebraic topology more deeply, you will see this pattern repeated over and over: new algebraic topology functors put to use in disproving existence of maps of various sorts.

Answer 2

There isn't really a single good way, just like there isn't a single good way to show that two topological spaces are non-homeomorphic, or that two groups are non-isomorphic. The best general tip, I think, would be to look for a property that deformation retractions conserve (or change in a predictable manner), and show that it is different for the two spaces.

In the case of a torus to the wedge of two circles, the fundamental groups are different, so one cannot be a deformation retraction of the other.

For the mobius strip, they do have isomorphic fundamental groups. But a deformation retraction is a homotopy, meaning of we take any loop in the mobius strip, sees where in the boundary the deformation retraction takes it, then map it back to the mobius strip by inclusion, we should still get the same element of the fundamental group. In this case, we clearly don't, as no loop contained in the boundary of the strip can be a generator for the fundamental group of the strip.

Or glue two mobius strips together along their edge to make a Klein bottle. If you deformation retract each of the mobius strips to their boundary, then the Klein bottle deformation retracts to a circle (the common boundary of the two strips). But a Klein bottle decidedly does not have the same fundamental group as a circle. So this is impossible.