4
$\begingroup$

In Topology from the Differentiable Viewpoint by Milnor the "Homotopy Lemma" is stated as follows

Homotopy Lemma: Let $f, g : M \to N$ be smoothly homotopic maps between manifolds of the same dimension, where $M$ is compact and without boundary. If $y \in N$ is a regular value for both $f$ and $g$, then $$\#f^{-1}(y) = \#g^{-1}(y) \ \ \ \text{(mod $2$)}$$

Now when proving this Milnor let's $F : M \times [0, 1] \to N$ be a smooth homotopy between $f$ and $g$. He then supposes that $y \in N$ is a regular value for $F$. I won't write out the details here , but he proceeds to show that $\# f^{-1}(y) = \# g^{-1}(y) \ \ (\text{mod} \ 2)$.

In the next part of the proof he supposes that $y \in N$ is not a regular value for $F$ and shows again that $\# f^{-1}(y) = \# g^{-1}(y) \ \ (\text{mod} \ 2)$ (using the result from the first part of the proof).

Now I'm assuming that Milnor must have proved something stronger, that for any point $y \in N$, $\#f^{-1}(y) = \# g^{-1}(y) \ \ (\text{mod} \ 2)$, because any point $y \in N$ is either a regular value for $F$ or it is not a regular value for $F$, and both cases would be covered by the proof above.

But then that leads me to question the following. Is $y \in N$ a regular value for both $f$ and $g$ if and only if $y \in N$ is a regular value for $F$? I'm not sure exactly how to prove this at the moment, or to provide a counterexample, but if the reverse direction holds, then the Homotopy Lemma would be proved. Does the reverse direction hold, do both directions hold, or does neither?

Finally is my assumption that Milnor proved something stronger correct? If so why hasn't Milnor stated this stronger result in the book instead of the weaker version. If my assumption is not correct, then how has Milnor actually proved the Homotopy Lemma without actually assuming $y \in N$ is a regular value for both $f$ and $g$?


Edit : Actually I just realized that in the second part of the proof $y$ is supposed to be not a regular value of $F$, but it is a regular value for $f$, otherwise $\# f^{-1}(y)$ wouldn't even be defined (in the proof on page 22). So the forward direction of my bolded question above doesn't seem to hold.

Also are there any assumptions that Milnor is assuming implicitly that I'm not aware of?

$\endgroup$
  • $\begingroup$ What if $F$ has no regular value at all ? $\endgroup$ – Nicolas Hemelsoet Oct 9 '17 at 20:59
  • $\begingroup$ Could you clarify the notation: what does $#$ mean here? $\endgroup$ – ಠ_ಠ Oct 9 '17 at 20:59
  • $\begingroup$ @ಠ_ಠ : # is the cardinal of a set $\endgroup$ – Nicolas Hemelsoet Oct 9 '17 at 20:59
  • $\begingroup$ @ಠ_ಠ If $f : M \to N$ is a smooth function between manifolds with $M$ compact and $y \in N$ is a regular value for $f$, then $\#$ is simply the cardinality of the set $\# f^{-1}(y)$ $\endgroup$ – Perturbative Oct 9 '17 at 21:01
  • $\begingroup$ This lemma certainly does not hold for values which are not regular for both maps. Does this answer your question? $\endgroup$ – Amitai Yuval Oct 9 '17 at 21:19
1
$\begingroup$

I was incorrect. In the proof Milnor does assume that $y \in N$ is a regular value for both $f$ and $g$, by mistake I didn't read the word "also" in the second line of the proof so I just assumed he was trying to prove something stronger using a theorem he hadn't proved before (and expected the reader to fill in the details).

What Milnor did was the following. He let $y \in N$ be regular values for both $f$ and $g$. Then in addition he first let $y$ be a regular value for $F$.

The whole reason he did this is because it's easier prove the Homotopy Lemma in the setting where we consider $y$ to be a regular value for both $f$, and $g$ and $y$ either a regular value for $F$ or not a regular value for $F$.

So firstly he proves that $\# f^{-1}(y) = \# g^{-1}(y) \ \ (\text{mod } 2)$, in the case that $y$ is a regular value for $F$, then he uses that result to prove $\# f^{-1}(y) = \# g^{-1}(y) \ \ (\text{mod } 2)$, in the case where $y$ is not a regular value for $F$.

Let $\Phi = \{y \in N \ | \ \text{y is a regular value for both $f$ and $g$}\}$, $\Gamma = \{y \in N \ | \ \text{y is a regular value for both $f$ and $g$ and $F$}\}$ and let $\Psi = \{ y \in N \ | \ \text{y is a regular value for both $f$ and $g$ and not $F$}\}$, then clearly $\Phi = \Gamma \cup \Psi$, so this proof covers all possible regular values for both $f$ and $g$.

Thus the proof makes sense.


To answer by bolded question, the backward direction doesn't seem to hold either.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.