General Topology – Homework: The Klein Bottle Retracts onto a Loop

Question

From the UCLA topology qualifying exam, Fall 2011:

Let $K$ be the Klein bottle. Show that there exist homotopically nontrivial simple closed curves $\gamma_1,\gamma_2$ such that $K$ retracts to $\gamma_1$, but does not retract to $\gamma_2$.

A candidate for $\gamma_2$ would be any of the curves that dissect $K$ into two Möbius bands, since the Möbius band cannot retract onto its boundary.
However I am completely stumped when it comes to finding $\gamma_1$.
The one thought I've had is that whatever $\gamma_1$ is, it cannot be a curve that partitions $K$ into two connected spaces, for then each side doesn't interact and for each side we'd be trying to retract a compact connected manifold onto its boundary, which is not possible.

Some hints or advice would be greatly appreciated.

Answer 1

We can consider the Klein Bottle $K$ to be the quotient space of the cylinder $C = S^1 \times I$ by the identification $(z,0) \sim (z^{-1},1)$. Here we consider $S^1$ as the unit circle in the complex plane. Let $L = 1 \times I$. I claim that $K$ retracts onto the loop $L'$ defined by $L$ under the quotient map.

Let $q : C \to K$ be the quotient map. The map $q$ satisfies the following universal property:

If $g : C \to Z$ is a continuous map from $C$ to any topological space $Z$ such that $a \sim b$ implies $g(a) = g(b)$ for all $a,b \in C$ then there exists a unique map $f : K \to Z$ such that $g = f \circ q$.

Define a map $g : C \to S^1$ as follows: First project onto $L$, then quotient to $L'$. This is a continuous map which is preserved under $\sim$, hence a continuous map $f : K \to L'$ exists and must be the identity on $L'$ by the universal property.