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 In this post, we’ll walk through the details which I learned from some unpublished notes of Greg Conner and David Fearnley.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

ineq1and if t\in[1/2,1], we have

ineq2Thus, 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<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.
  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

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One Response to The locally path-connected coreflection II

  1. Pingback: The locally path-connected coreflection III | Wild Topology

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