Is the order on connectivity classes total?

Question

Suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Such «relative connectivity» has a following property:

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Proof: Suppose $x_0, x_1 \in X$. Then $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now, we can say that topological spaces $X$ and $Y$ are connectively equivalent iff $X$ is $Y$-connected and $Y$ is $X$-connected. This equivalence relation divides all topological spaces onto connectivity classes.

There exists a natural order on connectivity classes: if $C_0$ and $C_1$ are connectivity classes, we say that $C_0 \leq C_1$ iff for any topological spaces $X \in C_0$ and $Y \in C_1$ we have that $X$ is $Y$-connected.

Connectivity classes have following properties:

There exists the smallest connectivity class and it is exactly the class of $1$-element spaces

Proof:

Suppose $X := \{x\}$ is $!$-element space, and $Y$ is an arbitrary topological space. Then $f: Y \to X$ such that $\forall y \in Y f(y) = x$ is the corresponding continuous function (no matter what $y_0$ and $y_1$ we take. Thus $X$ is $Y$-connected. All we need to prove that if $|Y| > 1$ then $Y$ is not $X$-connected is to consider two distinct elements of $Y$. Thus the class of all $1$-element spaces is the smallest connectivity class.

There exists the largest connectivity class and it is exactly the class of all disconnected spaces

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected.

So, we know, that there exists the largest connectivity class and it contains all disconnected spaces. To finish the proof we need to prove, that it does not contain any connected spaces:

Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Connectivity classes do not form a set

Proof can be found here

Now, my question is:

Is the order on connectivity classes a total order?

Or, equivalently:

Do there exist two topological spaces $X$ and $Y$ such that neither $X$ is $Y$-connected nor $Y$ is $X$-connected?

Answer 1

It is not a total order. Basically, if you take two connected spaces that are not path-connected for "different reasons", then they will probably be incomparable in the connectivity order. For instance, let $X$ be the topologist's sine curve and let $Y$ be the closed long line. Then every map $Y\to X$ is eventually constant, so $X$ is not $Y$-connected since it is not path-connected. On the other hand, every map $X\to Y$ has image that is separable and connected, so no such map can connect the endpoints of $Y$.

(To explain the importance of avoiding path-connected spaces, the path-connected spaces are the least connectivity class if you restrict to reasonably nice (say, normal) spaces with more than one point, as a consequence of Urysohn's lemma. See this answer for some related discussion.)

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Here's a rather different sort of counterexample. Let $X=\{a,b,c\}$ with the topology that a set is open iff it is $X$ or it does not contain $b$. Let $Y$ have the same underlying set but the opposite topology: the open sets of $Y$ are the closed sets of $X$. Then $X$ and $Y$ are incomparable in the connectivity order, since neither one can connect the points $a$ and $c$ in the other. (The quick way to see this is by thinking about their specialization orders: one looks like $\wedge$ and the other looks like $\vee$.)