What is the relation between the usual topology of $S^1$ and its subspace topology in Homeo$(S^{n+1})$?

Question

Let the set of all self homeomorphisms of $S^{2n+1}$ - $\operatorname{Homeo}(S^{2n+1})$, be given the compact open topology. Fix $a_0,\cdots,a_n\in\mathbb Z$ to be $n+1$ coprime integers. Let $S^1$ act on $S^{2n+1}$ as follows - $$\lambda\cdot(z_0,\cdots,z_n)=(\lambda^{a_0}z_0,\cdots,\lambda^{a_n}z_n)$$

$($The resulting quotient space is what is known as the weighted projective space $W\mathbb P(a_0,\cdots,a_n))$

The action is clearly faithful and hence we can think of $S^1$ as a sub-group of $\operatorname{Homeo}(S^{2n+1})$. We know that $S^1$ with its usual topology is a compact Lie group.

1. What can be said about $S^1$ as a subspace of $\operatorname{Homeo}(S^{2n+1})$? Will the subspace topology on $S^1$ be the same as its usual topology?

2. Will the topology change depending on the choice of $a_i$?

Thank you.

Answer 1

The induced map $S^1\to \operatorname{Homeo}(S^{2n+1})$ is a continuous injection from a compact space to a Hausdorff space, so it is a homeomorphism onto its image. That is, the subspace topology is the same as the usual topology of $S^1$.