5

It seems that I am being able to show that the dimension of a homeomorphic image of a topological space is equal to the dimension of the original space itself.

In what follows, I define the dimension of a topological space $X$ in accordance with Hartshorne: $$\dim X:= \sup \{n \in \mathbb N : \exists Z_0 \subsetneq \cdots \subsetneq Z_n \text{ s.t. } Z_j \subset X \text{ is closed and irreducible for all }j \in \{0, \cdots , n\}\}$$

Let $X$ and $Y$ be topological spaces and $f$ a homeomorphism from $X$ to $Y$ and let $r:= \dim X$. Then there exists a chain $$V_0 \subsetneq \cdots \subsetneq V_n$$ of closed irreducible subsets of $X$. $f$ being a closed map, each of $f(V_0), \cdots , f(V_n)$ must also be closed and clearly $$f(V_0) \subset \cdots \subset f(V_n).$$ Moreover, for each $j \in \{1, \cdots , n\}$, considering $z_j \in V_j \setminus V_{j-1}$ gives us from the injectivity of $f$, $f(z_j) \in f(V_j) \setminus f(V_{j-1})$, so that all the containments mentioned in the above chain are proper.

Finally, we note that each $f(V_j)$ is irreducible simply because the image of an irreducible set under a continuous map is also irreducible: indeed for any irreducible closed set $Z$ of $X$, if $f(Z) = Y_1 \cup Y_2$ for two nonempty proper closed subsets $Y_j$ of $Y$, then $$Z = f^{-1}(f(Z)) = f^{-1}(Y_1 \cup Y_2) = f^{-1}(Y_1) \cup f^{-1}(Y_2),$$ with each $f^{-1}(Y_j)$ closed (in $X$), non-empty and proper by the continuity, surjectivity and injectivity of $f$ respectively, thereby contradicting the irreducibility of $Z$.

Thus, $\dim Y \geq r = \dim X$ and the opposite inequality follows by switching the roles of $X$ and $Y$, since the inverse of a homeomorphism is also a homeomorphism. It seems that homeomorphisms are preserving the (topological) dimension of a topological space, contrary to the counterexample provided here: Do homeomorphic sets have to be in the same dimension?

I would be really obliged if someone could please find the fallacy in the above argument.

Edit: I changed the title after Moishe Kohan's clarification.

AK12N1
  • 157
  • 2
    The linked post is utter nonsense, just ignore it. Incidentally, your notion of dimension is coming from algebraic geometry and topologists never use it, for instance, since ${\mathbb R}^n$ with classical topology contains only two types of closed irreducible subsets (empty set and singletons). – Moishe Kohan Jan 22 '21 at 03:10
  • Oh, I see, thank you so much. It is interesting indeed how a notion based entirely on topology makes such little appearance in the subject itself. – AK12N1 Jan 22 '21 at 03:12
  • 2
    Topologists have their own notions of dimension, see for instance here. – Moishe Kohan Jan 22 '21 at 03:13
  • I see, thank you. So just for a final confirmation, the above proof works then right? – AK12N1 Jan 22 '21 at 03:17
  • 2
    Yes, of course. But if you are doing algebraic geometry, the notion of Zariski-homeomorphisms is not very useful. See, however, this question. – Moishe Kohan Jan 22 '21 at 03:18
  • 1
    @AK12N1: The algebraic-geometry definition of dimension is not completely divorced from the topology one: It gives the expected answer for $\mathbb{A}^n$; it behaves as expected under products; it agrees with the dimension of the tangent space at nice points; and so on. The topological concept of dimension is that an $n$-dimensional space should be topologically similar to $\mathbb{R}^n$, though, and the (non-Hausdorff) Zariski topology is very different from the usual one on $\mathbb{R}^n$; $\mathbb{A}^n(\overline{\mathbb{F}_p})$ certainly doesn't look like $\mathbb{R}^n$ topologicaly. – anomaly Jan 22 '21 at 03:20

0 Answers0