Connectifications? - MathOverflow

Connectifications?

2011-06-01T11:41:48Z

Like many of my questions, this question is actually aimed at $p$-adic analysis. One of the main obstacles in doing analysis $p$-adically ist that the $\mathbb{Q}_p$ is totally disconnected.

From previous answers and reading I learned that one tries to circumvent these problems and the result are things like "rigid analytic spaces" or "Berkovich spaces".

Then I recently thought. Ok $\mathbb{Q}_p$ if it is not connected why not do what often is done when some property does not hold:

incomplete -> take the completion

not compact -> compactify

not algebraically closed -> take algebraic closure.

So why not just connectify $\mathbb{Q}_p$? Now my questions is twofold.

Searching on the for connectify or connectification I only got a few results, which only seem to be from point set topology without applications in other areas of mathematics. Why is that so? Is connectification somehow bad behaved, or does it not exist in general? By connectification I would understand something like for algebraic closure or completion or the stone-cech compactification which can all be defined by a universal property. Does this make sense at all to define connectification like that.
Coming back to the case $\mathbb{Q}_p$, I guess that one also wants the connectification of $\mathbb{Q}_p$ to be a field. Is this maybe not satisfied, or are there other reasons not to consider the connectification of $\mathbb{Q}_p$.

Answer by Henno Brandsma for Connectifications?

2011-06-01T12:05:19Z

They do not always exist (I believe the Sorgenfrey line does not have one, e.g.), and if they exist they are not very well-behaved.

This might be a relevant paper

Answer by Qiaochu Yuan for Connectifications?

2011-06-01T12:09:52Z

The two-point discrete space already doesn't have a (universal) connectification, in the sense that two points don't have a coproduct in the category of connected spaces. If $X$ were such a coproduct, then given any pair of points in a connected space $C$ there would have to be a unique compatible map $X \to C$. But letting $C = [0, 1]$ and the points be $0, 1$ we can easily twist any such map by a homeomorphism $C \to C$ fixing $0$ and $1$.

Answer by Todd Trimble for Connectifications?

2011-06-01T13:44:23Z

After seeing wood's last comment (comment #2 under his question), I've decided to add a few words (a bit too many for a comment) which hopefully make clear the force of Qiaochu's answer.

Generally speaking, the categorical meaning of "completion" refers to taking a left adjoint of a full inclusion of categories; in our situation we are considering the full subcategory

$$\text{Conn} \hookrightarrow \text{Top}$$

from connected spaces to general topological spaces. Examples where such completions exist are: the inclusion of complete metric spaces into the category of metric spaces and continuous maps with Lipschitz constant 1 (Cauchy completion), the inclusion of compact Hausdorff spaces into the category of all spaces (Stone-Cech compactification), and the inclusion of fields into the category of integral domains and injective ring maps (field of fractions construction).

(There's a bit of fine print here: sometimes one also demands that the unit of the adjunction, here the universal map of an object to its completion, be injective. For example, the inclusion of abelian groups into the category of groups does have a left adjoint (the abelianization), but this isn't injective. Similarly, to get the map from a space to its Stone-Cech compactification to be injective, one should really consider the inclusion of compact Hausdorff spaces in the category of completely regular spaces. Sometimes the suffix -ization or -ification is used in cases where the unit is not injective.)

The salient point behind Qiaochu's answer is that a left adjoint, if it exists, must preserve coproducts (or in fact colimits generally). Now, supposing that the left adjoint to the inclusion $\text{Conn} \to \text{Top}$ exists, it would first of all take a one-point space to a one-point space (the proof is easy), and it would take a coproduct of two one-point spaces in $\text{Top}$, viz. a two-point discrete space, to a coproduct of two one-point spaces in $\text{Conn}$. But Qiaochu's example shows this cannot possibly exist.

The only remedy that I can think of in this situation is to change things up a bit, in a way that I don't think will be at all useful to the OP. There are for example situations where an algebraic structure on a space forces it to be connected (and then some), where one can construct the corresponding free algebraic structures to get a left adjoint to the (non-full) forgetful functor mapping to $\text{Top}$. The most obvious example might be to consider spaces equipped with a contraction: consider spaces $X$ equipped with a basepoint $x_0: 1 \to X$ and with an action $\alpha: [0, 1] \times X \to X$ of the multiplicative monoid $[0, 1]$, such that $\alpha(0, x) = x_0$ for all $x \in X$. Here the free algebra on a general space $Y$ is just the cone $CY$ with the obvious algebraic structure. But this isn't likely to be useful to the OP.