Which properties exactly do homeomorphisms preserve?

I'm trying to clarify which properties homeomorphisms preserve and why.

It was noted in this answer, that homeomorphisms preserve topology and therefore any property that can be defined in a general topological space. I am looking for a more precise characterization of such properties and a proof (or at least an explanation, if it is obvious from something that I missed) of why this is true.

Solution 1:

Example 1: Metrizability. Let $(X,\mathcal T_X)$ be a topological space. The linked answer is concerned with metrizability, i.e. the property that one can define a metric $d_X:X\times X\to\mathbb R$ such that this metric exactly reproduces $\mathcal T_X$ as the set of open sets.

Imagine $(X,\mathcal T_X)$ is a metrizable topological space, and let $(Y, \mathcal T_Y)$ be homeomorphic to $(X,\mathcal T_X)$, with a homeomorphism $f:X\to Y$. Can we prove that $(Y, \mathcal T_Y)$ is a metrizable space? Easy: define $d_Y(y_1, y_2)=d_X(f^{-1}(y_1), f^{-1}(y_2))$, then prove $d_Y$ is a metric on $(Y,\mathcal T_Y)$ that exactly reproduces $\mathcal T_Y$ as the set of open sets.

Example 2: Compactness. Let $(X, T_X)$ be a topological space. We will say that it is compact if, for every family $(U_i)_{i\in I}$ of subsets $U_i\in\mathcal T_X$ such that $X=\bigcup_{i\in I}U_i$ there is a finite subfamily $U_{i_1}, U_{i_2},\ldots, U_{i_n}$ (with $i_1,i_2,\ldots, i_n\in I$) such that already $X=\bigcup_{j=1}^n U_{i_j}$.

Imagine now $(X,\mathcal T_X)$ is a compact topological space, and let $(Y, \mathcal T_Y)$ be homeomorphic to $(X, \mathcal T_X)$ with a homeomorphism $f:X\to Y$. Can we prove that $(Y, \mathcal T_Y)$ is a compact space? Easy: let $(V_i)_{i\in I}$ be a family of subsets from $\mathcal T_Y$ such that $Y=\bigcup_{i\in I}V_i$. Map those sets using $f^{-1}$ to get the family $U_i=f^{-1}(V_i)$. As $f$ is a homeomorphism, it is continuous, so $U_i\in\mathcal T_X$, and because it is a bijection, we can prove that $X=\bigcup_{i\in I}U_i$. Now, as $(X,\mathcal T_X)$ is compact, a finite subfamily of $U_i$'s already has $X$ as their union: $X=\bigcup_{j=1}^nU_{i_j}$ for some $i_1, i_2,\ldots, i_n\in I$. Now, apply $f$ to prove that $V_{i_j}=f(U_{i_j})$ have the same property in $(Y,\mathcal T_Y)$, i.e. $Y=\bigcup_{j=1}^n V_{i_j}$

Example 3: Hausdorff property. Let $(X, T_X)$ be a topological space. We will say that it is Hausforff if, for every two elements $x_1, x_2\in X, x_1\ne x_2$ there are two elements $U_1, U_2\in\mathcal T_X$ such that $x_1\in U_1, x_2\in U_2$ and $U_1\cap U_2=\emptyset$.

Imagine now $(X,\mathcal T_X)$ is a Hausdorff topological space, and let $(Y, \mathcal T_Y)$ be homeomorphic to $(X, \mathcal T_X)$ with a homeomorphism $f:X\to Y$. Can we prove that $(Y, \mathcal T_Y)$ is a Hausdorff space? Easy: let $y_1, y_2\in Y, y_1\ne y_2$. Map them using $f^{-1}$: $x_1=f^{-1}(y_1), x_2=f^{-1}(y_2)$. Obviously $x_1\ne x_2$ and $x_1,x_2\in X$, so because $(X,\mathcal T_X)$ is Hausdorff, there are two sets $U_1, U_2\in\mathcal T_X$ such that $x_1\in U_1, x_2\in U_2$ and $U_1\cap U_2=\emptyset$. Map those two sets back using $f$: let $V_1=f(U_1), V_2=f(U_2)$. Because $f^{-1}$ is continuous, $V_1, V2\in\mathcal T_Y$. Also, $y_1=f(x_1)\in f(U_1)=V_1$ and $y_2\in V_2$. Finally, as $f$ is "one to one", $V_1\cap V_2=\emptyset$ too. As this is valid for any chosen $y_1, y_2\in Y, y_1\ne y_2$, the space $(Y, \mathcal T_Y)$ is Hausdorff too.

Generalization: Whatever other property, defined only using the set $X$ and the collection $\mathcal T_X$, you can define on a topological space $(X, \mathcal T_X)$, you can prove that the same property holds on any other topological space $(Y,\mathcal T_Y)$ homeomorphic to $(X, \mathcal T_X)$. Namely, pick the homeomorphism $f:X\to Y$, and then work through the definition of the property, using $f$ and $f^{-1}$, back and forth, to map elements and subsets. The complexity of the proof will be of about the same order of magnitude as the complexity of the definition of the property, anyways.

Disclaimer: Obviously I've left the part "Generalization" without a proof. I don't know of a good reference where this has been proven formally: hope someone else may come up with one. Note the reference will need to include the definition for the language that we need to use to express those "topological properties" (so that we can possibly mount a proof based on the induction on the length of the definition).

Solution 2:

Ordinary mathematics is typically taken to occur in ZFC, where the "basic notion" is the elementhood relation $\in$. Given any $a$ and $b$, it makes sense to ask whether $a \in b$ or whether $a = b$.

This leads to some undesirable results. For example, we would like two spaces $X$ and $Y$ which are homeomorphic to be totally indistinguishable. However, this is not the case - for example, $\mathbb{R}$ and $(0, 1)$ are distinguished by whether $2$ is a point in the space. It also allows us to ask silly questions, like "Is $1$ an element of $\cos$?" or "What is the intersection of $e$ and $\mathbb{C}$?", which shouldn't even have answers to begin with.

To get around this, we can reformulate the axioms of set theory purely in terms of functions. Here, we have a primitive notion of sets (denoted with capital letters). Given sets $A$ and $B$, we have a primitive notion of functions between $A$ and $B$ (written with lowercase letters $f : A \to B$), and we have a primitive notion of function composition (given $f : A \to B$, $g : B \to C$, we have $g \circ f : A \to C$). We also are only permitted to write $a = b$ when $a$ and $b$ are functions with the same domain and codomain - there is no notion of equality of sets.

The first axioms we use are those of category theory. We assume, for all $A$, the existence of an identity arrow $1_A$ which is a left and right identity for composition. We assume associativity of composition.

We then posit the existence of a 1-element set $1$ satisfying the axiom $\forall A \exists! f : A \to 1$. In the usual manner of Category Theory, we then posit the existence of Cartesian Products and equalisers. We also introduce the idea of power sets as "power objects", giving us a topos.

Finally, we introduce some notation. The "elements of $A$" are defined to be the functions $1 \to A$, and we define the notation $x :\in A$ to mean $x : 1 \to A$. Given $x :\in A$ and $f : A \to B$, we define $f(x) :\in B$ to be $f \circ x$.

We then add the axiom of function extensionality, which states $\forall A \forall B \forall f, g : A \to B, (\forall x :\in A, f(x) = g(x)) \to g = f$.

Once we've defined "elements", we can define "$f : A \to B$ is surjective" to mean $\forall b :\in B \exists a :\in A, f(a) = b$. We then state and assume the axiom of choice: for all surjective $f : A \to B$, there is some $g : B \to A$ such that $f \circ g = 1_B$.

At this point, we have a theory known as ETCS. We can add the axiom scheme of replacement to this theory (formulated appropriately in the language of category theory) to get a theory ETCSR such that the following statement is true:

Thm. For any sentence $\phi$ in the language of ETCSR, $ETCSR \vdash \phi$ if and only if ZFC proves that $\phi$ holds in the category of sets. Thus, ZFC and ETCSR are "equivalent".

Now, we can formalise the notion of "isomorphism invariance of truth". Note that in the next two paragraphs, when we discuss "equality", we are discussing syntactic equality of variable names in the metatheory.

Consider object variables $O_i$, where $i$ ranges from $1$ to $n$, and consider object variables $P_j$ and $Q_j$ where $j$ ranges from $n + 1$ to $n + k$. Let $P_j = Q_j = O_j$ whenever $1 \leq j \leq n$, so that $P_j$ and $Q_j$ are defined for all $j$ from $1$ to $n + k$.

Consider arrow variables $f_i : O_{d_i} \to O_{c_i}$, where $i$ ranges from $1$ to $m$, and where for all such $i$, we have $1 \leq d_i, c_i \leq n$. Consider arrow variables $g_j : P_{d_j} \to P_{c_j}$ and $h_j : Q_{d_j} \to Q_{c_j}$, where $j$ ranges from $m + 1$ to $m + w$ and where $1 \leq d_j, c_j \leq n + k$ for all such $j$. Define $g_j = f_j = h_j$ for all $1 \leq j \leq m$.

Finally, consider arrow variables $v_j : P_j \to Q_j$ for $n + 1 \leq j \leq n + k$. Define $v_i = 1_{O_i}$ for $1 \leq i \leq n$.

Let $\phi$ be any statement whose free variables consist only of the $O$s, $P$s, $f$s, and $g$s (and not of the $Q$s, $h$s, or $v$s). Then using only the axioms of identity arrows and associativity of composition, we can prove the universal closure of the following:

Suppose that for all $j$ such that $n + 1 \leq j \leq n + k$, $v_j$ is an isomorphism. Further suppose that for all $i$ such that $m + 1 \leq i \leq m + k$, we have that $h_i \circ v_{d_i} = v_{c_i} \circ g_i$. Then $\phi \iff \phi[P \mapsto Q, g \mapsto h]$.

This statement is known as the "isomorphism invariance of truth", and it can be proven by a simple structural induction on $\phi$.

Here are some applications of the isomorphism invariance of truth:

Consider a terminal object $A$ and an isomorphism $f : A \to B$. Then $B$ is a terminal object.

Proof: We take $P_1 = A$, $Q_1 = B$, and $v_1 = f$ and apply isomorphism invariance of truth.

Consider maps $f, g : A \to B$ and an equaliser $e : X \to A$. Now consider an isomorphism $h : B \to C$. Then $e$ is also the equaliser of $h \circ f, h \circ g : A \to C$.

Proof: We can rephrase this as follows:

Given arrows $f, g : A \to B$ and arrows $f', g' : A \to X$, if there is an isomorphism $h : B \to X$ such that $f' = h \circ f$ and $g' = h \circ f$, then $e$ is the equaliser of $f$ and $g$ iff $e$ is the equaliser of $g'$ and $h'$. This follows immediately from isomorphism invariance of truth. Apply the above to the case $f' = h \circ f$ and $g' = h \circ g$.

Now, let's turn out attention to topological spaces. Notice that a topological space consists of some set $X$ and some subset $\tau_X \subseteq PX$ satisfying some properties. Note that a homeomorphism $f : X \to Y$ is an isomorphism; it therefore induces an isomorphism $PX \to PY$ which in turn induces an isomorphism between the subsets $\in_X \subseteq X \times PX$ and $\in_Y \subseteq Y \times PY$ that "plays nicely" with the induced isomorphism $X \times PX \to Y \times PY$. Because $f$ is a homeomorphism, it will also induce an isomorphism $\tau_X \to \tau_Y$ which plays nicely with the other isomorphisms. This gives us enough for isomorphism invariance.

Therefore, any structural property at all - that is, any property that can be stated purely in terms of the category of sets - is automatically homeomorphism invariant. This includes everything we could possibly care about - metrizability, compactness, 2nd-countability, being a manifold of dimension $n$, the various separation axioms, and more. Algebraic invariants like homotopy groups, homology, and cohomology are also preserved, as are properties of the categories of presheaves and sheaves on the space.

The only things homeomorphisms can fail to preserve in ZFC are those which cannot be phrased in terms of the category of sets - things like "Is $2$ an element of the underlying space?" But if we work in ETCSR (or a weaker theory), we can't even phrase these questions to begin with.