Is the top hemisphere of $\mathbb{S}^1$ a retract? Consider $\mathbb{S}^1$, centered at $0$ in $\mathbb{R^2}$. Consider the top hemisphere $\mathbb{H} = \{ (x,y) \in \mathbb{S}^1 \mid y \geq 0 \} $.
Is there a continuous retraction from $\mathbb{S}^1$ to $\mathbb{H}$?
Attempt: Yes, because reflection across the x-axis is continuous?

More generally,  the map $(x_1,\dots,x_n)\to (x_1,\dots,|x_n|)$ is a retraction of $S^{n-1}$ onto its upper half. As was noted in comments, this is not a deformation retraction; there isn't one since the hemisphere is contractible while the sphere is not.
