A smooth map of a sphere to itself is homotopic to a map with isolated fixed points

Question

Let $v:S^k\to S^k$ be a smooth map of a sphere into itself. Such a map possibly can have nonisolated fixed points (e.g. the identity map of $S^k$). Can we always homotope $v$ to a smooth map $S^k\to S^k$ which has only isolated (hence, finitely many) fixed points?

Actually this is asserted in Lemma 7.4 of Milnor's book Singular Points of Complex Hypersurfaces. It is written in the proof that: After perturbing $v$ slightly we may assume that the fixed points of $v$ are all isolated. But I can't understand how perturbing $v$ "slightly" gives a map with isolated fixed points.

Here is one idea: Consider the embedding $V:S^k\to S^k\times S^k$, $V(x)=(x,v(x))$. The fixed points of $v$ correspond to the intersection $\operatorname{image}(V)\cap \Delta$, where $\Delta=\{(x,x)\in S^k\times S^k:x\in S^k\}$ is the diagonal. Suppose $V$ is homotopic to a map $W$ which is transverse to $\Delta$ (such a map $W$ always exists). Then $\operatorname{image}(W)\cap \Delta$ must be finite by dimensional results, and if $w$ is the composition $S^k \xrightarrow{W} S^k\times S^k \xrightarrow{\text{proj}_2} S^k$, then $v$ is homotopic to $w$. Now does $w$ have only finitely many fixed points? (This would be true if $W=(\text{id},w)$, but in the general case..)

Is this argument correct? Still I cannot understand what "slightly" means in the book.

P.S. In fact Lemma 7.4 assumes that $v(x)\neq -x$ for all $x\in M$, where $M$ is a compact codimension zero submanifold of $S^k$ with boundary, containing no fixed points of $v$ in the interior. Maybe the homotopy of $v$ should be taken so that it is fixed on $M$.

Answer 1

Your argument is good, but let me elaborate on a few points:

Now does $w$ have only finitely many fixed points?

The answer is no. For example, suppose $k = 1$ and that $v : S^1 \to S^1$ is the identity map. It is easy to find a perturbation $v'$ of $v$ which has no fixed points. Then the map $W = (v', \mathrm{id}) : S^1 \to S^1 \times S^1$ is transverse to $\Delta$, but $w = \mathrm{proj}_2 \circ W$ is the identity map, which certainly does not have isolated fixed points.

One way to remedy this issue is to constrain the perturbation $W$ to be of the form $(\mathrm{id}, w)$. Now the question to ask is whether $W$ can always be chosen to be transverse to the $\Delta$. The answer is yes and can be seen via the transversality theorem as follows. Embed $S^k$ in $\mathbb{R}^{k + 1}$ in the usual way and write $\pi : \mathbb{R}^{k + 1} - \{0\} \to S^k$ for the usual projection. For a small ball $B_\varepsilon \subseteq \mathbb{R}^{k + 1}$ centered at the origin, consider the family of maps $$ B_\varepsilon \times S^k \to S^k \times S^k, \quad (s, x) \mapsto (x, \pi(v(x) + s)). $$ This map is transverse to $\Delta$; indeed, at each point the image of its differential is the tangent space to the second $S^k$-factor. Thus, by the transversality theorem, for a generic parameter $s \in B_\varepsilon$, the map $$ W_s(x) = (x, \pi(v(x) + s)) $$ is transverse to $\Delta$. Since $W_0 = V$, we are done by taking $W = W_s$ for generic $s$.

Note that we have essentially reproven what is sometimes called the "transversality homotopy theorem" (the statement that a smooth map can always be homotoped to be transverse to a submanifold), but with an added constraint on the resulting map (namely, that its first component is the identity).

Still I cannot understand what "slightly" means in the book.

Intuitively, this means we should be able to choose $w$ to be an arbitrarily small perturbation of $v$. This certainly seems possible. Indeed, the above argument produces a family of maps $w_s = \mathrm{proj}_2 \circ W_s$ such that $w_s$ has isolated fixed points for generic $s \in B_\varepsilon$. The precise meaning of "generic" is that $s$ can be chosen away from a measure zero set of $B_\varepsilon$ (cf. Sard's theorem). In particular, we can choose $s$ to be arbitrarily close to $0$. Morally speaking, this should imply $w_s$ is an arbitrarily small perturbation of $v$. However, we have not made precise the meaning of a perturbation being "arbitrarily small."

There are various ways to make this more precise. The simplest way would be to choose a metric on $S^k$, which makes the space of smooth maps $C^\infty(S^k, S^k)$ into a metric space via the $C^0$-metric $$ \operatorname*{dist}(f, g) = \mathrm{sup}_{x \in S^k}\operatorname*{dist}(f(x), g(x)). $$ Then a formal statement would be that $w$ can be chosen to be arbitrarily close to $v$ in the $C^0$-metric. Note we have essentially proven this already; it remains to convince yourself that the map $s \mapsto w_s$ is a continuous map from $B_\varepsilon$ to $C^\infty(S^k, S^k)$ (equipped with the $C^0$-metric).

However, being close in the $C^0$-metric is a somewhat weak condition. For instance, the derivative of the perturbation $w$ may behave poorly. A good exercise might be to define a $C^1$-metric (or more generally, a $C^k$-metric) on the space $C^\infty(S^k, S^k)$ and to show that $w$ can be made close to $v$ in the $C^1$-metric. Such a metric would be far from canonical (indeed, this is the case for the $C^0$-metric), so perhaps it would be easier to define instead a $C^1$-topology. Then the desired statement would be that $w$ can be chosen within any open neighborhood of $v$ in the $C^1$-topology. A reference for this material would be Chapter 2 of Hirsch's Differential Topology.