Homotopy equivalence of compositions involving Hopf map

Question

Recall that one definition of the Hopf map is given by $\eta:S^3 \rightarrow S^2:(z_0,z_1) \mapsto z_0/z_1$, where $S^3$ is the unit sphere in $\mathbb{C}^2$ and $S^2 = \mathbb{C} \cup \infty$.

Here is an exercise related to this:

Show that $k\eta \simeq \eta k^2$, where $k$ and $k^2$ denote maps of degree $k$ and $k^2$, respectively.

The notion of degree that I'm used to is the degree of a map $f:S^n \rightarrow S^n$ with $n > 0$. It doesn't seem like $\eta$, for instance, is a map of this form, but I think that we can still make sense of its degree by thinking of the degree of $\eta$ restricted to $S^2 \subseteq S^3$.

We know that if $\text{deg}(f) = \text{deg}(g)$, then $f \simeq g$. We also know that $\text{deg}(fg) = \text{deg}(f) \cdot \text{deg}(g)$. Lastly, we know that if $f$ is not surjective, then its degree is $0$. We will use all of these facts to our advantage below.

Hence, we can show that $k \eta \simeq \eta k^2$ by showing that $\text{deg}(k \eta) = \deg(\eta k^2)$. We have $\deg(k \eta) = \deg(k) \cdot \deg(\eta) = k \cdot \deg(\eta)$ and $\deg(\eta k^2) = \deg(\eta) \cdot \deg(k^2) = k^2 \cdot \deg(\eta)$. Thus, to show the desired equality $\text{deg}(k \eta) = \deg(\eta k^2)$, it suffices to show that $\deg(\eta) = 0$.

Is it really true that the degree of the Hopf map is equal to $0$ when restricted to $S^2$? Such is commented on the post here, but I can't see why this true. One way we could show this is by showing that the Hopf map is not surjective when restricted to $S^2$ -- is this true? Can we see this simply from how the Hopf map is defined above?

Thanks!

The degree, as you're working with it, is only defined for self maps of a sphere, not for maps between spheres of different dimensions. Hint: the map $(z_0,z_1)\rightarrow (z_0^k, z_1)$ is a degree $k$ map of $S^3$, and the map $(z_0/z_1)\mapsto (z_0/z_1)^k$ is a degree $k$ map of $S^3$. — Jason DeVito - on hiatus, Apr 18 '22 at 03:49
Notice also that, if the degree of a map from $S^3$ to $S^2$ were additive, the degree of any map would be zero (think about the constante map), hence it does nota distinguish honotopy classes. — user17786, Apr 18 '22 at 07:49
@JasonDevito Are you suggesting that we build an explicit homotopy here? — michiganbiker898, Apr 22 '22 at 16:04
Not necessarily. Given the degree $k$ map on $S^3$, can you form a degree $k^2$ map on $S^3$ of a similar form? — Jason DeVito - on hiatus, Apr 22 '22 at 16:07
@JasonDevito I was thinking that the composition $k \eta$ sends $(z_0,z_1)$ to $(z_0/z_1)^k$, while $\eta k^2$ sends $(z_0,z_1)$ to $z_0^{k^2}/z_1$. — michiganbiker898, Apr 22 '22 at 16:16
So, you can either show those two maps are homotopic, or come up with a different degree $k^2$ map of $S^3$. I recommend the latter. — Jason DeVito - on hiatus, Apr 22 '22 at 16:31
@JasonDevito Hatcher's text explains how to construct a map from $S^n \rightarrow S^n$ of arbitrary degree (Example 2.31). Is that construction what you had in mind? — michiganbiker898, Apr 23 '22 at 06:04
No, I am recommending you model the construction of a degree $k^2$ map if $S^3$ similarly as to the degree $k$ one I wrote. As a further hint, the map $(z_1,z_2)\mapsto (z_1,z_2^k)$ is also degree $k$. — Jason DeVito - on hiatus, Apr 23 '22 at 12:11
@JasonDevito Aah, I see. In that case, $k \eta$ sends $(z_0,z_1)$ to $(z_0/z_1)^k$, while $\eta k^2$ sends $(z_0,z_1)$ to $z_0^k / z_1^k$. Is that what you were thinking? — michiganbiker898, Apr 23 '22 at 18:41
Yep! No need to write down homotopies. The two maps are equal! — Jason DeVito - on hiatus, Apr 23 '22 at 18:54
Feel free to write up an answer to your own question, if you want. — Jason DeVito - on hiatus, Apr 23 '22 at 19:35

Homotopy equivalence of compositions involving Hopf map

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