Which of these are homotopy equivalent? $S^1, \mathbb{R}, \{*\}$

Question

Which of these spaces are homotopy equivalent: $S^1, \mathbb{R}, \{*\}$?

I found a homotopy equivalence between $\mathbb{R}$ and the one point space $\{*\}$, so they are homotopy equivalent.

The sphere and $\mathbb{R}$ are not homotopy-equivalent because the $\pi_1(S^1) \cong \mathbb{Z}$ whereas $\pi_1(\mathbb{R})$ is trivial (since we just proved that $\mathbb{R}$ is contractible). Is there a nicer way of seeing this without using the fundamental group?

Is $S^n$ for $n>1$ homotopy equivalent to $\mathbb{R}$?

Answer 1

You've asked several questions at once:

Which of these spaces are homotopy equivalent: $S^1, \mathbb{R}, \{*\}$?

You've pretty much figured out the answer to this question already. You know $\mathbb{R}$ and the one-point space are homotopy equivalent, and $S^1$ and $\mathbb{R}$ are not homotopy equivalent. Homotopy equivalence is an equivalence relation, so you don't need to check if $S^1$ is homotopy equivalent to the one-point space. You know that it's not, because otherwise $S^1$ and $\mathbb{R}$ would be homotopy equivalent by transitivity.

Is there a nicer way of seeing this without using the fundamental group?

This is pretty much why fundamental groups and other homotopy invariants exist; in order to distinguish spaces.

Is $S^n$ for $n>1$ homotopy equivalent to $\mathbb{R}$?

No, but I'm not sure how to answer this question without going further into algebraic topology. As I alluded in the comments, there are higher homotopy groups $\pi_k(X)$ which are also homotopy invariants. It can be shown $\pi_k(S^n) = 0$ if $1 \leq k < n$, and $\pi_n(S^n) = \mathbb{Z}$. Also, $\pi_k(X) = 0$ for all $k \geq 1$ if $X$ is contractible. So this tells you $S^n$ and $\mathbb{R}$ are not homotopy equivalent.