# Find the homotopy classes of based maps between two given spaces.

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.

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 $$f_*(\pi_1(\mathbb{R}P^n)) = 0$$.
• 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. Jul 20, 2021 at 4:43
• @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 $f$. Jul 20, 2021 at 5:00