Find the homotopy classes of maps $[S^n,S^1]$ and $[\mathbb RP^n, S^1],$ for $n \geq 2.$

I know the following theorem in the context of covering spaces.

Let $p : E \longrightarrow B$ be a covering map; let $p(e_0) = b_0.$ Let $f : Y \longrightarrow B$ be a continuous map, either $f(y_0) = b_0.$ Suppose $Y$ is path connected and locally path connected. The map $f$ can be lifted to a map $\widetilde {f} : Y \longrightarrow E$ such that $\widetilde {f} (y_0)) = e_0$ if and only if $$f_{*} (\pi_1 (Y,y_0) \subseteq p_* (\pi_1 (E,e_0)).$$

Furthermore, if such a lifting exists, it is unique.

With the help of the above theorem how do I deduce the required homotopy classes of maps?

I know that $\pi_1 (S^n) = 0,$ for $n \geq 2$ and $S^n$ is cleary path connected and locally path connected. Therefore with the help of the above theorem we can say that any map $f : S^n \longrightarrow S^1$ factors through $\mathbb R$ via the following composition of maps $$S^n \xrightarrow {\widetilde {f}} \mathbb R \xrightarrow {p} S^1$$ where $\widetilde {f}$ is the unique lifting of $f$ from $S^n$ to $\mathbb R$ and $p : \mathbb R \longrightarrow S^1$ is the covering map. Now how do I proceed? Does the contractibility of $\mathbb R$ play any role here?

For the second one I didn't able to conclude that $f_*(\pi_1(\mathbb R P^n)) = 0 = p_* (\pi_1(\mathbb R)).$ I know that $\pi_1(\mathbb R P^n) = \mathbb Z/2\mathbb Z.$ But how does it imply that $f_*(\pi_1(\mathbb R P^n)) = 0\ $? Which is needed to apply the above theorem.

Any help in this regard would be much appreciated. Thanks for your time.


1 Answer 1


Yes, the contractibility of $\mathbb{R}$ plays a central role here. For you have just shown that all maps $\mathbb{S}^n \to \mathbb{S}^1$, $n \geq 2$, factor through $\mathbb{R}$. This means that there is at most one map $\mathbb{S}^n \to \mathbb{S}^1$ up to homotopy, since if we have $f, g : \mathbb{S}^n \to \mathbb{S}^1$, then $\tilde{f} \simeq \tilde{g}$ because $\mathbb{R}$ is contractible, and hence $f = p \circ \tilde{f} \simeq p \circ \tilde{g} = g$.

In the second case, we know that for all $u \in \pi_1(\mathbb{R}P^n)$, $u \cdot u = e$. Therefore, we have $f_*(u) \cdot f_*(u) = 0$ in $\pi(\mathbb{S}^1)$. Since the latter is isomorphic to $\mathbb{Z}$, we see that it must be the case that $f_*(u) = e$ for all $u$, and hence $f_*(\pi_1(\mathbb{R}P^n)) = 0$.

  • $\begingroup$ Why does $\mathbb R$ contractible imply $\widetilde {f} \simeq \widetilde {g}\ $? Is it true that if the range of a continuous map is contractible then the map is nullhomotopic? For the second part I got the point because there doesn't exist any non-trivial homomorphism $\varphi : \mathbb Z/2\mathbb Z \longrightarrow \mathbb Z.$ In fact we can argue along the lines that $\text {ord} (\varphi (a))\ \vert\ \text {ord} (a),$ for any $a \in \mathbb Z/2\mathbb Z.$ But all the non-identity element of $\mathbb Z$ is of infinite order and hence the result follows. $\endgroup$ Jul 20, 2021 at 4:43
  • 1
    $\begingroup$ @AntonioClaire If the codomain of a continuous map is contractible then the map is nullhomotopic. This is because a contractible space is a space $Y$, such that there is a point $y$ such that $1_Y \simeq const_y$. So for all $f : X \to Y$, we have $f = 1_Y \circ f \simeq const_y \circ f$. And $const_y \circ f$ is independent of $f$. $\endgroup$ Jul 20, 2021 at 5:00
  • $\begingroup$ Oh! Very nice explanation. I am totally convinced. Thanks for your help. $\endgroup$ Jul 20, 2021 at 5:13
  • $\begingroup$ @AntonioClaire You're welcome. $\endgroup$ Jul 20, 2021 at 5:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .