## The locally path-connected coreflection II

In the last post, I discussed how to efficiently change the topology of a space $X$ in order to obtain a locally path-connected space $lpc(X)$ without changing the homotopy or (co)homology groups of the space in question. This is a handy thing to have hanging from your tool belt but there are some reasonable concerns that come along with this type of construction.

Say $X$ is metrizable. Must the $lpc$-coreflection also be metrizable? It turns out the answer is yes but that we might lose separability along the way. In this post, we’ll walk through the details which I learned from some unpublished notes of Greg Conner and David Fearnley.

Theorem: If $X$ is path-connected and metrizable, then there is a metric inducing the topology of $lpc(X)$ such that the identity function $id:lpc(X)\to X$ is distance non-increasing.

Proof. Suppose $X$ is a space whose topology is induced by a metric $d$. Define a distance function $\rho$ on $lpc(X)$ as follows: For any path $\alpha:[0,1]\to X$ and $t\in[0,1]$, let $$\ell_t(\alpha)=d(\alpha(0),\alpha(t))+d(\alpha(t),\alpha(1)).$$

Observe that $d(\alpha(0),\alpha(1))\leq\ell_t(\alpha)$ for any $t$ by the triangle inequality.

Now let $$\ell(\alpha)=\sup\{\ell_t(\alpha)|t\in[0,1]\}$$

For points $a,b\in X$, we define our metric as $$\rho(a,b)=\inf\{\ell(\alpha)|\alpha\text{ is a path from }a\text{ to }b\}$$

Since $d(a,b)\leq\ell(\alpha)$ for any path $\alpha$ from $a$ to $b$, we get that $d(a,b)\leq \rho(a,b)$ showing that the identity $lpc(X)\to X$ is non-increasing.

We should still check that $\rho$ is actually a metric which induces the topology of $lpc(X)$.

Some notation first: If $\alpha,\beta$ are paths in $X$ such that $\alpha(1)=\beta(0)$,  then $\alpha^{-}(t)=\alpha(1-t)$ denotes the reverse of $\alpha$ and $\alpha\cdot\beta$ denotes the usual concatenation of paths $$\alpha\cdot\beta(t)=\begin{cases} \alpha(2t) & 0\leq t\leq 1/2\\ \beta(2t-1) & 1/2\leq t\leq 1 \end{cases}$$

Notice that $\ell(\alpha)=\ell(\alpha^{-})$ and given $t\in[0,1/2]$, we have

and if $t\in[1/2,1]$, we have

Thus, in general, $\ell(\alpha\cdot\beta)\leq\ell(\alpha)+\ell(\beta)$. Now we can check that $\rho$ is a metric.

1. If $a=b$, then we may take $\alpha$ to be the constant path at this point. Then $\ell(\alpha)=0$ showing $\rho(a,b)=0$. Conversely, if $a\neq b$, consider any path $\alpha:[0,1]\to X$ from $a$ to $b$. Find $0 such that $\alpha(t)\notin\{a,b\}$. Then $0. Since $\alpha$ was arbitrary, we have $\rho(a,b)>0$.
2. Symmetry $\rho(a,b)=\rho(b,a)$ is clear since for every path $\alpha$ from $a$ to $b$, there is a unique reverse path $\alpha^{-}$ from $b$ to $a$ with $\ell(\alpha)=\ell(\alpha^{-})$.
3. Suppose $a,b,c\in X$. Let $\alpha$ be any path from $a$ to $b$ and $\beta$ be any path from $b$ to $c$. Then there is a path $\alpha\cdot\beta$ is a path from $a$ to $c$ such that $\ell(\alpha\cdot\beta)\leq\ell(\alpha)+\ell(\beta)$. Therefore $\rho(a,c)\leq\rho(a,b)+\rho(b,c)$ finishing the proof that $\rho$ is a metric.

The metric topology induced by $\rho$ is finer than the topology of $lpc(X)$: Suppose $U$ is an open set in $X$ (with the topology induced by $d$) and $C$ is some path component of $U$. Let $x\in C$. Find an $\epsilon$-ball such that $B_{d}(x,\epsilon)\subseteq U$. We claim that $B_{\rho}(x,\epsilon)\subseteq C$: if $y\in B_{\rho}(x,\epsilon)$, then $\rho(x,y)<\epsilon$ so there is a path $\alpha:[0,1]\to X$ from $x$ to $y$ such that $\ell(\alpha)<\epsilon$. Since $d(x,\alpha(t))\leq\ell_t(\alpha)\leq\ell(\alpha)<\epsilon$ for all $t\in[0,1]$, we conclude that $\alpha(t)\in B_{d}(x,\epsilon)\subseteq U$ for all $t$. Since $\alpha$ has image in $U$, we must have $\alpha(1)=y\in C$, proving the claim.

The topology of $lpc(X)$ is finer than the metric topology induced by $\rho$: For the other direction, suppose $B_{\rho}(x,\epsilon)$ is an $\epsilon$-ball with respect to $\rho$. Pick a point $y\in B_{\rho}(x,\epsilon)$ and let $\delta=\epsilon-\rho(x,y)$. We claim that the path-component of $y$ in $B_{d}(y,\delta/4)$ is contained in $B_{\rho}(x,\epsilon)$. Let $\alpha$ be a path in $B_{d}(y,\delta/4)$ such that $\alpha(0)=y$. It suffices to check that $z=\alpha(1)\in B_{\rho}(x,\epsilon)$. Notice that $\ell_{t}(\alpha)=d(y,\alpha(t))+d(\alpha(t),z)< \frac{\delta}{4}+\frac{\delta}{2}= \frac{3\delta}{4}$ for all $t\in[0,1].$ Thus $\ell(\alpha)\leq \frac{3\delta}{4}$ showing that $\rho(y,z)<\delta$. We now have $$\rho(x,z)\leq\rho(x,y)+\rho(y,z)<(\epsilon-\delta)+\delta=\epsilon$$proving the claim.$\square$

Example: One thing to be wary of is that $lpc(X)$ can fail to be separable even if $X$ is a compact metric space. For instance, let $A$ be a Cantor set in $[1,2]$. Then we can use the construction of generalized wedges of circles in the previous post to construct the planar set $X=C_A$ which is a compact metric space (and certainly separable). This is basically a wedge of circles where the circles are parameterized by a Cantor set. But $lpc(X)$ is an uncountable wedge of circles (with a metric topology – not the CW topology – at the joining point) and this is not separable. The general problem here seems to be that there might be open sets of $X$ which have uncountably many path-components!

For any given space $Y$, $\pi_0(Y)$ will denote the set of path-components of $Y$.

Theorem: Let $X$ be a metric space. Then $lpc(X)$ is separable if and only if $X$ is separable and $\pi_0(U)$ is countable for every open set $U\subseteq X$.

Proof. If $lpc(X)$ is separable, then since the identity function $lpc(X)\to X$ is continuous and surjective, $X$ is separable as the continuous image of a separable space. Now pick a countable dense set $A\subset lpc(X)$ and let $U$ be a non-empty open set in $X$. Now $\pi_0(U)$ is the set of path-components of $U$. If $C\in\pi_0(U)$, then $C$ is open in $lpc(X)$ and thus there is a point $a\in A\cap C$. This gives a surjection from a subset of $A$ onto $\pi_0(U)$ showing that $\pi_0(U)$ is countable.

For the converse, if $X$ is a separable metric space then it has a countable basis $\mathscr{B}$. Furthermore, we assume $\pi_0(B)$ is countable for every set $B\in\mathscr{B}$. Let $\mathscr{C}=\bigcup_{B\in\mathscr{B}}\pi_0(B)$ be the collection of all path-components of the basic open sets. Then $\mathscr{C}$ is countable. If $C$ is the path-component of $x$ in an open set $U$ of $X$, then there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U.$ Now if $D$ is the path-component of $x$ in $B$, then $x\in D\subseteq C$ where $D\in\mathscr{C}$. This shows $\mathscr{C}$ forms a countable basis for the topology of $lpc(X)$. Since $lpc(X)$ is metrizable (by our above work), it is also separable.$\square$