Can someone explain why amoebas are closed?

Question

From Wikipedia,

Claim: any amoeba is closed.

I attempted to use the preimage of closed under continuous map is closed but I realized that my skill with complex functions was too feeble, so I found a proof instead

http://www.sciencesmaths-paris.fr/upload/Contenu/Notes%20et%20resumes%20cours/Notes_Cours_Mikhalkin/lecture-notes1.pdf

Can someone use simple language or a series of simple propositions to explain what the proof is trying to do? If I were to attempt to use the standard proof: the preimage of closed set under continuous map is closed, what are the necessary ingredients contained in this proof?

Thanks!

Answer 1

It is true that the continuous preimage of a closed set is closed. This is just a general fact of topology:

• $A$ is closed iff (by definition) its complement $A^c$ is open.

• $f$ is continuous iff (by definition) the $f$-preimage of any open set is open.

• The complement of the preimage is the preimage of the complement. (Exercise)

However, this isn't what you want: you want that the image of a closed set is closed (since you're trying to go from $V$ to $Log(V)$, and you know the former is closed). This is in general false: for example, consider the map $arctan(x)$ as a function $\mathbb{R}\rightarrow\mathbb{R}$. Is the $f$-image of the closed set $\mathbb{R}$ a closed set?

A map which sends closed sets to closed sets is called (unsurprisingly) closed, and in general continuous maps need not be closed and closed maps need not be continuous.

This seems to rule out an easy proof . . .

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. . . so we're going to have to be more clever. What other properties does $Log$ have?

Well, the crucial observation is:

The map $Log: (\mathbb{C}^\times)^N\rightarrow\mathbb{R}^N$ is proper (that is, preimages of compact sets are compact).

This isn't hard to show: a compact set in either $(\mathbb{C}^\times)^N$ or $\mathbb{R}^N$ is just a closed and bounded set, so it's enough to show that the $Log$-preimage of a bounded set is bounded since $Log$ is continuous.

OK, so what? Well, it turns out that if $f:X\rightarrow Y$ is a continuous proper map and $Y$ is locally compact (in fact, we only need the weaker condition that $Y$ be "compactly generated," but that's a bit more technical), then $f$ is closed! I won't prove this here, since we don't need it; I'll just prove that any continuous proper map between two locally compact metric spaces is closed.

• Suppose $f:X\rightarrow Y$ is a continuous proper map between two locally compact metric spaces, and $A\subseteq X$ is some closed set. Let $y\in\overline{A}$ (where "$\overline{\cdot}$" denotes "closure"); then for each natural number $i$, let $U_i$ be a compact subset of $Y$ containing $y$ with diameter $<2^{-i}$, and let $W_i=f^{-1}(U_i)\cap A$.

• Since $y$ is on the boundary of $f(A)$, $W_i\not=\emptyset$ for any $i$. So pick $x_i\in W_i$.

• Since $f$ is proper, $f^{-1}(U_i)$ is compact, and so the sequence $(x_i)_{i\in\mathbb{N}}$ has a convergent subsequence. Without loss of generality, suppose the whole sequence converges, and call its limit point $z$.

• Since $A$ is closed, $z\in A$. But do you see how to show now that $f(z)=y$?