In a exercise I need to find a retract $r:S^1 \cup[0,2] \times\{0\} \to S^1 \subset\mathbb R^2$.

Question

Hi I am studying algebraic topology and I have trouble with finding retracts. In a exercise I need to find a retract $r:S^1 \cup[0,2] \times\{0\} \to S^1 \subset\mathbb R^2$. I had the following in mind: $$r(x) = \frac{x}{\|x\|}$$ Here $S^1\subset \mathbb R^2$ is the unit circle. But we run into trouble with the point $(0,0)\in S^1 \cup[0,2] \times\{0\} $. I have not been able to find a way to evade this problem point. Can someone help me or give me a hint on how to find the right retract?

Answer 1

Let's take a look at the function you have in mind: for any $x \in S^{1},$ $\lVert x \rVert$ is $1,$ so it's just the identity. Then, for any $x$ on the segment $[0, 2] \times \{0\}$ not equal to $(0, 0),$ we see that $\frac{x}{\lVert x \rVert}$ is just the point $(1, 0).$ Unfortunately, $\frac{x}{\lVert x \rVert}$ is not defined at $(0, 0).$

However, this gives us an idea of what our desired function should do: it should be the identity on $S^{1},$ and send all of $[0, 2] \times \{0\}$ to $(1, 0).$

So, let's define our function this way: let $$r(x) = \begin{cases} \text{id}_{S^{1}} & \text{if} \; x \in S^{1} \\ (1, 0) & \text{if} \; x \in [0, 2] \times \{0\}.\end{cases}$$

We need to check this is well-defined: on the intersection $S^{1} \cap ([0, 2] \times \{0\}) = \{(1, 0)\},$ we see that $\text{id}_{S^{1}}$ and the constant function sending everything to $(1, 0)$ agree. So, we are good here.

Now, we need to show that this $r$ is continuous.

Note that $r|_{S^{1}}$ is the identity, hence is continuous. Similarly, $r|_{[0, 2] \times \{0\}}$ is the constant map sending everything to $(1, 0)$, hence is also continuous.

Finally, note that $S^{1}, [0, 2] \times \{0\}$ are two closed subspaces of $S^{1} \cup ([0, 2] \times \{0\}),$ whose union is the whole space. So, by the Pasting Lemma, $r(x)$ is continuous.

Since $r|_{S^{1}} = \text{id}_{S^{1}}$, $r$ is a retract to $S^{1}$, as desired.