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, 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.
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