If $X$ is a topological space, one can naturally view the set $\pi\_0(X)$ of path-components of $X$ as a quotient space of $X$ by collapsing each path-component to a point by a quotient map $q:X\to \pi\_0(X)$. Of course, $q$ is a homeomorphism if and only if $X$ is totally path-disconnected. I'd like to know of criteria that ensure $\pi\_0(X)$ is totally path-disconnected.


Interestingly, there are lots of spaces $X$ for which $\pi\_0(X)$ is not totally path-disconnected. For instance, if $X=[0,1]\times [0,1]$ has the lexicographical ordering and is given the resulting order topology, then $q:X\to \pi\_0(X)\cong [0,1]$ is just the projection onto the first coordinate. In fact, a little known gem due to Douglas Harris is that for every space $Y$, one can construct a paracompact Hausdorff space $S(Y)$ such that $\pi\_0(S(Y))\cong Y$.


D. Harris, [Every space is a path-component space](https://msp.org/pjm/1980/91-1/pjm-v91-n1-p09-s.pdf), Pacific J. Math. 91 no. 1 (1980) 95-104.


However, the ordered square and spaces $S(Y)$ are not metrizable...


One must also be wary of separation axioms since if $X$ is the closed topologists sine curve, $\pi\_0(X)$ is the two-point Sierpinski space, which is not $T\_1$ and therefore not totally path-disconnected.


**Question:** If $X$ is a metric space and $\pi\_0(X)$ is $T\_1$, must $\pi\_0(X)$ be totally path-disconnected?


*Note: I'm interested in other variations of the question too where $X$ is separable or perhaps Polish and/or $\pi\_0(X)$ is Hausdorff.*


**Update:** I'm happy to say that this post motivated collaboration between Taras Banakh and myself. This MO-inspired work resulted in the following publication:


T.Banakh, J.Brazas, *Realizing spaces as path-component spaces*, Fundamenta Mathematicae 248 (2020) 79-89. [DOI: 10.4064/fm529-12-2018](https://www.impan.pl/en/publishing-house/journals-and-series/fundamenta-mathematicae/all/248/1/113082/realizing-spaces-as-path-component-spaces)



Paths in path component spaces
There exists the following metric counterpart of the Harris result:


**Theorem** ([Banakh, Vovk, Wojcik](https://arxiv.org/pdf/0901.0236.pdf)): Each metric space $X$ can be (canonically) identified with the space $\pi\_0(\circledast(X))$ of path-components of some complete metric space $\circledast(X)$.


The space $\circledast(X)$ is a closed subspace of the complete oriented graph with vertices in the set $X$.


The above theorem follows from Theorem 7.3 and Proposition 7.4 of [this paper](https://arxiv.org/pdf/0901.0236.pdf) published in Fund. Math. 212 (2011), 145-173.




---


More generally, the Theorem holds for weakly first-countable spaces containing no countable connected subspaces. To formulate the general and more precise version of the Theorem, I need to recall some definition and results from [this paper](https://arxiv.org/pdf/0901.0236.pdf) of Banakh, Vovk and Wojcik, which will be cited as [BVW].
So, I thank Jeremy Brazas for opportunity to advertise this my paper [BVW] :)


Unfortunately, writing the paper [BVW] we did not know about the paper of Harris. After looking at the Harris' paper, I discovered that his construction of the space $S(X)$ (with $\pi\_0(S(X))\cong X$) is similar to our construction of $\circledast(X)$ with $\pi\_0(\circledast(X))\cong X$. But our space $\circledast(X)$ always is complete metric. As a result, our construction works only for premetric (= weakly first-countable) spaces.




---


Let us recall that a topological space $X$ is *weakly first-countable* if to each point $x\in X$ one can assign a decreasing sequence $(B\_n(x))\_{n\in\mathbb N}$ of subsets containing $x$ such that a subset $U\subset X$ is open if and only if for each point $x\in U$ there exists $n\in\mathbb N$ such that $B\_n(x)\subset U$.


By Proposition 3.7 in [BVW], a topological space $X$ is weakly first-countable if and only if the topology of $X$ is generated by a premetric $p$.


A *premetric* on a set $X$ is any function $p:X\times X\to[0,\infty)$ such that $p(x,x)=0$ for all $x\in X$. 
The topology generated by a premetric $p$ on $X$ consists of all sets $U\subset X$ such that for every $x\in X$ there exists $\varepsilon>0$ such that the $\varepsilon$-ball $B(x;\varepsilon):=\{y\in X:p(x,y)<\varepsilon\}$ is contained in $U$.


A *premetric space* is a pair $(X,p)$ consisting of a set $X$ and a premetric $p$.


To each premetric space $X$ we shall assign some special complete metric space $\circledast(X)$, called the *cobweb* of the premetric space $X$.


The cobweb space is a closed subset of the complete oriented graph $\Gamma X$ over the set $X$. The complete oriented graph $\Gamma X$ is the set $X\cup\{(x,y,t)\in X\times X\times(0,1):x\ne y\}$ endowed with the path metric $d$ (in which every oriented edge $[x,y]=\{x,y\}\cup\{(x,y,t):0<t<1\}$ has length 1.


For a premetric space $X$ endowed with a premetric $p$ the *cobweb* $\circledast(X)$ is defined as the closed subset 
$$\circledast(X):=X\cup\{(x,y,t)\in\Gamma(X):t\le 1-p(y,x)\}$$of the graph $\Gamma X$.


The *compression map* $\pi\_X:\circledast(X)\to X$ assigns to each point $a\in \circledast(X)$ the point $a$ if $a\in X$ and the point $x$ if $a=(x,y,t)$ for some $x,y\in X$ and $t\in(0,1)$. 


The following theorem follows from Theorem 7.3 and Proposition 7.4 of [BVW].


**General Theorem** (Banakh, Vovk, Wojcik) *Let $X$ be a premetric space. Then:* The cobweb space $\circledast(X)$ is complete metric space of dimension $\le 1$.
* The compression map $\pi\_X:\circledast(X)\to X$ is a continuous quotient surjection with path-connected fibers $\pi\_X^{-1}(x)$, $x\in X$.
* The space $\circledast(X)$ is connected if and only if $X$ is connected.
* If $X$ contains no countable connected subspaces, then the fibers of the compression map $\pi\_X$ coincide with the path-components of $\circledast(X)$ and also with the separablewise components of $\circledast(X)$;*


We recall that the *separablewise component* of a point $x$ in a topological space $X$ is the union of all separable connected subspaces of $X$ that contain $x$. It is clear that the path-component of $x$ is contained in the separablewise component of $X$.


Since the topology of a weakly first-countable space is generated by a premetric, 
the General Theorem implies the following corollary answering the question of
 Jeremy Brazas in (strong) negative.


**Corollary 1.** *Each weakly first-countable space is homeomorphic to the space $\pi\_0(M)$ of path-components of some complete metric space $M$.*


Taking for $X$ any connected Hausdorff space, we obtaine the following surprising 


**Example.** *There exists a connected Polish space $P$ such that $P$ has countably many path-connected components and each path-connected component is closed in $P$.*




---


It seems that for separable space $X$ the space $\pi\_0(X)$ also can be homeomorphic to $[0,1]$. A possible counteexample can look as follows:


Let $$C=\big\{\sum\_{i=1}^\infty
\frac{x\_i}{3^i}:(x\_i)\_{i\in\mathbb N}\in\{0,2\}^{\mathbb N}\big\}$$ be the standard Cantor set in the unit interval $I:=[0,1]$.


Let $\mathcal J$ be the set of connected components of the complement $I\setminus C$. 
It is clear that each set $J\in\mathcal J$ is an open interval of lentgth $\frac1{3^k}$ for some $k\ge 1$.


For $n\in\{0,1\}$ let $\mathcal C\_n$ be the subset of $\mathcal C$ consisting of intervals of length $\frac1{3^{2i+n}}$ for some $i\ge 0$.
Let $J\_n=\bigcup\mathcal J\_n$.


Now consider the following compact subset $$X:=(C\times[0,1])\cup(\{0\}\times J\_0)\cup(\{1\}\times J\_1).$$
I hope that the space $\pi\_0(X)$ of path-components of $X$ is homeomorphic to $[0,1]$.


