The locally path-connected coreflection (Part 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 In this post, we’ll walk through the details which I learned from some unpublished notes of Greg Conner and David Fearnley.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.

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 $0<t<1$ such that $\alpha(t)\notin\{a,b\}$ . Then $0<d(a,b)\leq \ell_t(\alpha)\leq\ell(\alpha)$ . Since $\alpha$ was arbitrary, we have $\rho(a,b)>0$ .
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 $\ell(\alpha)=\ell(\alpha^{-})$ .
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. $\square$

For one more post on this see: Locally path-connected coreflection (Part III)

5 Responses to The locally path-connected coreflection (Part II)

Pingback: The locally path-connected coreflection III | Wild Topology
Pingback: The locally path-connected coreflection | Wild Topology
David Roberts says:

September 9, 2019 at 1:44 am

Some more here (copied and pasted from the post as I see it):

$$\ell_t(\alpha)=d(\alpha(0),\alpha(t))+d(\alpha(t),\alpha(1)).$$

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

$$\rho(a,b)=\inf\{\ell(\alpha)|\alpha\text{ is a path from }a\text{ to }b\}$$

$$ \alpha\cdot\beta(t)=\begin{cases} \alpha(2t) & 0\leq t\leq 1/2\\ \beta(2t-1) & 1/2\leq t\leq 1 \end{cases}$$

LikeLike

- Jeremy Brazas says:
  
  September 9, 2019 at 10:54 pm
  
  Thanks again David!
  
  LikeLike
  
Pingback: The locally path-connected coreflection (Part III) | Wild Topology