Homotopically Hausdorff Spaces II

In my first post on homotopically Hausdorff spaces, I wrote about the property which describes the existence of loops that can be deformed into arbitrarily small neighborhoods but which are not actually null-homotopic, i.e. can’t be deformed all the way back to that point. In this post, we’ll offer up different viewpoints on this property based on an approach taken from a recent paper:

Jeremy Brazas, Hanspeter Fischer, Test map characterizations of local properties of fundamental groups. Preprint. 2017. Click here for the arXiv paper.

In particular, we’ll discuss the following characterization of the homotopically Hausdorff property.

Theorem 1: For a first countable space X, the following are equivalent:

  1. X is homotopically Hausdorff,
  2. Every map every map f:\mathbb{HA}\to X from the harmonic archipelago induces the trivial homomorphism f_{\#}:\pi_1(\mathbb{HA},b_0)\to\pi_1(X,f(b_0)) on \pi_1.
  3. For every map g:\mathbb{H}\to X such that g_{\#}([\ell_n])=1\in \pi_1(X,f(b_0)) for every n\geq 1, then f_{\#}([\ell_1\cdot\ell_2\cdots])=1.

The harmonic archipelago \mathbb{HA}. Notice the Hawaiian earring \mathbb{H} is a subspace and every loop based at the one wild point b_0 may be deformed over finitely many hills to lie within an arbitrary neighborhood of b_0.

Condition 2. in the theorem clarifies the notion that the harmonic archipelago really is the prototypical non-homotopically Hausdorff space since maps from it detect the same failure.

Condition 3. suggests that the homotopically Hausdorff property should be thought of as a closure property of the trivial subgroup.

Just as a reminder, here is the definition of the property we’re focusing on.

Definition: Given a path-connected space X and basepoint x_0, we considered the subgroup

\pi(\alpha,U)=\{[\alpha\cdot\gamma\cdot\alpha^{-}]|\gamma([0,1])\subseteq U\}\leq\pi_1(X,x_0)

where \alpha is a path starting at x_0 and U is an open neighborhood of \alpha(1). We say X is homotopically Hausdorff if for every path \alpha:[0,1]\to X with \alpha(0)=x_0, we have \bigcap_{U\in\mathcal{T}_{\alpha(1)}}\pi(\alpha,U)=1 where \mathcal{T}_x denotes the set of all open sets in X containing x.

Infinite Concatenations of Paths and Homotopies

We’re going to use the Hawaiian earring as a kind of “test space” so let’s recall its construction. If C_n is the circle of radius \frac{1}{n} centered at \left(\frac{1}{n},0\right), then \mathbb{H}=\bigcup_{n\geq 1}C_n is the usual Hawaiian earring space with basepoint b_0=(1,0). Let \ell_n:[0,1]\to\mathbb{H} be the loop which traverses C_n once in the counterclockwise direction. The homotopy classes [\ell_n], n\geq 1 freely generate the subgroup F=\langle [\ell_n]|n\geq 1\rangle.

Definition: A sequence \{\alpha_n\}_{n\geq 1} of paths \alpha_n:[0,1]\to X is null at a point x\in X if for every open neighborhood U of x, there is an N such that \alpha_n([0,1])\subseteq U for all n\geq 1, equivalently if \{\alpha_n\}_{n\geq 1} converges to the constant path at x.

Given a sequence of paths \{\alpha_n\}_{n\geq 1} satisfying \alpha_n(1)=\alpha_{n+1}(0) and which is null at x\in X, we may define the infinite concatenation to be the path \alpha=\prod_{n=1}^{\infty}\alpha_n to be the path defined to be \alpha_n on the interval \left[1-\frac{1}{n},1-\frac{1}{n+1}\right] and \alpha(1)=x.

Sometimes, we may expand the notation as

\displaystyle\prod_{n=1}^{\infty}\alpha_n=\alpha_1\cdot\alpha_2\cdot\alpha_3\cdots

Infinite Concatenation

Example: An important infinite concatenation for this post will be the loop

\ell_{\infty}=\prod_{n=1}^{\infty}\ell_n=\ell_{1}\cdot\ell_{2}\cdot\ell_{3}\cdots

that winds once around each hoop C_n of \mathbb{H}.

Warning: Notice here that we’re only considering infinite concatenations or “products” of loops – not homotopy classes of loops. Indeed, this operation is well defined for paths but the notion of “infinite product” of homotopy classes is not well defined in all fundamental groups.

Remark 2: What we are allowed to do with these infinite products is reparameterize them. This allows us to treat them like infinite sums and products in Calculus:

\prod_{n=1}^{\infty}\alpha_n\simeq \left(\prod_{n=1}^{m}\alpha_n\right)\cdot\left(\prod_{n=m+1}^{\infty}\alpha_n\right)

and so

\left[\prod_{n=1}^{\infty}\alpha_n\right]= \left[\prod_{n=1}^{m}\alpha_n\right]\left[\prod_{n=m+1}^{\infty}\alpha_n\right]

in the fundamental groupoid for any m

Proposition 3: A sequence of loops \{\alpha\}_{n\geq 1} based at x is null at x if and only if there is a map f:(\mathbb{H},b_0)\to (X,x) such that f\circ\ell_n=\alpha_n.

Proof. The key here is to observe that a function f:(\mathbb{H},b_0)\to (X,x) is continuous if and only if f|_{C_n} is continuous for each n and if for every neighborhood U of x, f maps all but finitely many of the circles C_n into U. The latter condition is clearly equivalent to the sequence of loops f\circ\ell_n being null at x. \square

Lemma 4: Let \{\alpha\}_{n\geq 1} be a null sequence of paths in X such that \alpha_n(0)=x for all n. Then the infinite concatenation

\prod_{n=1}^{\infty}(\alpha_{n}\cdot\alpha_{n}^{-})=(\alpha_1\cdot\alpha_{1}^{-})\cdot(\alpha_2\cdot\alpha_{2}^{-})\cdot(\alpha_3\cdot\alpha_{3}^{-})\cdots

is a null-homotopic loop.

Proof. Recall that for any path \alpha with \alpha(0)=x, we can contract \alpha\cdot\alpha^{-} to the constant path at x by a homotopy contracting the loop back along its own image.

At height t, the homotopy h(s,t) pictured is first \alpha|_{[0,t]}, constant in the black region, and then the reverse of \alpha|_{[0,t]}.


We construct a null-homotopy H:[0,1]\times[0,1]\to X of \prod_{n=1}^{\infty}(\alpha_{n}\cdot\alpha_{n}^{-}) by creating an infinite concatenation of the individual contractions h_n of \alpha_n\cdot\alpha_{n}^{-}. It will looks something like this:

Infinite concatenation of null-homotopies.

where H is defined as h_n on \left[1-\frac{1}{n},1-\frac{1}{n+1}\right]\times [0,1]. You can use the pasting lemma to verify continuity at every point except those on the right vertical wall. To verify continuity of H on the right edge recall that \{\alpha\}_{n\geq 1} is null at x. This means that given any open neighborhood U of x, there is an N such that \alpha_n has image in U for all n\geq N. But h_n has image in \alpha_n([0,1]) for each n. Therefore, H\left(\left[1-\frac{1}{N},1\right]\times [0,1]\right)\subseteq U. We conclude that there is an open set V containing \{1\}\times [0,1] which is mapped into U by h.

This verifies the continuity of H. \square

Functorality of the Obstruction

Lemma 5: Let f:X\to Y be a map, \alpha:[0,1]\to X be a path, and U be an open neighborhood of f(\alpha(1)). Then f_{\#}(\pi(\alpha,f^{-1}(U))\leq\pi(f\circ \alpha,U).

Proof. If \gamma is a loop in f^{-1}(U), then [\alpha\cdot\gamma\cdot\alpha^{-}] is a generic element of \pi(\alpha,f^{-1}(U)). Since f\circ \gamma has image in U, it follows that

f_{\#}([\alpha\cdot\gamma\cdot\alpha^{-}])=[(f\circ\alpha)\cdot(f\circ\gamma)\cdot(f\circ \alpha)^{-}]\in\pi(f\circ \alpha,U). \square

Corollary 6: Let f:X\to Y be a map, \alpha:[0,1]\to X be a path from x_0 to x, and set f(x_0)=y_0 and f(x)=y. Then

f_{\#}\left(\bigcap_{V\in\mathcal{T}_x}\pi(\alpha,V)\right)\leq\bigcap_{U\in\mathcal{T}_y}\pi(f\circ\alpha,U)

as subgroups of \pi_1(Y,y_0).

Proof. Suppose g\in\pi(\alpha,V) for all V\in\mathcal{T}_x and pick any U\in\mathcal{T}_y. Then g\in\pi(\alpha,f^{-1}(U)) and by Lemma 5, we have f_{\#}(g)\in\pi(f\circ\alpha,U). Thus f_{\#}(g)\in\bigcap_{U\in\mathcal{T}_y}\pi(f\circ\alpha,U). \square

Interpretation: Corollary 6 can be thought of as saying that the “obstruction” subgroups \bigcap_{U\in\mathcal{T}_x}\pi(\alpha,U) which detect the failure of the homotopically Hausdorff property are functorial since continuous maps induce homomorphisms that always map obstruction subgroups into obstruction subgroups.

Proof of Theorem 1

(1. \Rightarrow 2.)

In the harmonic archipelago \mathbb{HA} every loop based at b_0 may be continuously deformed within an arbitrary neighborhood U of the basepoint b_0. Thus if \alpha:[0,1]\to\mathbb{HA} is the constant path at the wild point b_0, then \pi(\alpha,U)=\pi_1(\mathbb{HA},b_0) for every open neighborhood U of b_0. Hence \bigcap_{U\in\mathcal{T}_{b_0}}\pi(\alpha,U)=\pi_1(\mathbb{HA},b_0).

Now suppose f:\mathbb{HA}\to X is a map such that the induced homomorphism f_{\#}:\pi_1(\mathbb{HA},b_0)\to\pi_1(X,f(b_0)) is not the trivial homomorphism, then by Corollary 6, we have

1\neq f_{\#}(\pi_1(\mathbb{HA},b_0))=f_{\#}\left(\bigcap_{V\in\mathcal{T}_{b_0}}\pi(\alpha,V)\right)\leq\bigcap_{U\in\mathcal{T}_{f(b_0)}}\pi(f\circ\alpha,U).

Now f\circ \alpha is a constant path at f(b_0) such that \bigcap_{U\in\mathcal{T}_{f(b_0)}}\pi(f\circ\alpha,U)\neq 1, which means X cannot be homotopically Hausdorff.

Note: this direction of Theorem 1 doesn’t actually require first countability.

(2. \Rightarrow 3.)

The main fact that we need is that if j:\mathbb{H}\to\mathbb{HA} is the inclusion map, then j_{\#}([\ell_{\infty}])\neq 1 in \pi_1(\mathbb{HA},b_0). I give a simple explicit proof of this fact in this post. A quick reminder of how this is done: compactness of the unit disk means that a null-homotopy of j\circ\ell_{\infty} can only intersect finitely many of the hills of \mathbb{HA}. So if E_n is the interior of the n-th hill, then j\circ\ell_{\infty} is null-homotopic in \mathbb{H}\cup\bigcup_{1\leq n\leq N}E_n for some N but this is impossible since \ell_{\infty} winds around the circle retracts C_n, n>N in a non-trivial way.

We prove the contrapositive. Suppose 3. does not hold. Then there exists a map g:\mathbb{H}\to X such that g_{\#}([\ell_n])=1 for all n\geq 1 and g_{\#}([\ell_{\infty}])\neq 1.

For each n\geq 1, the loop g\circ(\ell_n\cdot\ell_{n+1}^{-}):S^1\to X is null-homotopic loop in X and therefore extends to a map on the unit disk. Since each of the holes in \mathbb{H} can be extended to “large” disks, g extends to a map f:\mathbb{HA}\to X such that f\circ j=g. So we have [j\circ\ell_{\infty}]\neq 1 and f_{\#}([j\circ\ell_{\infty}])=g_{\#}([\ell_{\infty}])\neq 1. Therefore f_{\#}:\pi_1(\mathbb{HA},b_0)\to\pi_1(X,f(b_0)) is not the trivial homomorphism.

Note: This part of the proof does not require first countability either.

(3. \Rightarrow 2.)

For this direction of Theorem 1, we do need the assumption that X is homotopically Hausdorff. We prove the contrapositive.

Suppose that X is first countable and that X fails to be homotopically Hausdorff.  Then there exists a path \alpha:[0,1]\to X from x_0 to x and a loop \gamma based at x such that

1\neq[\alpha\cdot\gamma\cdot\alpha^{-}]\in\bigcap_{U\in\mathcal{T}_x}\pi(\alpha,U)\leq\pi_1(X,x_0).

Notice that if we conjugate by [\alpha]^{-1}, then we see that 1\neq [\gamma]\in\pi_1(X,x).

Let U_1\supset U_2\supset U_3\supset... be a countable neighborhood base at x. Then

1\neq[\alpha\cdot\gamma\cdot\alpha^{-}]\in\bigcap_{n\geq 1}\pi(\alpha,U_n)\leq\pi_1(X,x_0).

Hence, for each n\geq 1, there is a loop \gamma_n:[0,1]\to U_n based at x such that [\alpha\cdot\gamma\cdot\alpha^{-}]=[\alpha\cdot\gamma_n\cdot\alpha^{-}]. In particular, [\gamma]=[\gamma_n]  in \pi_1(X,x) for each n\geq 1.

By construction, the sequence of loops \{\gamma_n\}_{n\geq 1} is null at x. Therefore, the sequence \{\gamma_n\cdot\gamma_{n+1}^{-}\}_{n\geq 1} of loops is also null at x. Using Proposition 3, we put this sequence together to construct a continuous function g:(\mathbb{H},b_0)\to (X,x) defined by g\circ\ell_n=\gamma_n\cdot\gamma_{n+1}^{-}.

Notice that g_{\#}([\ell_n])=[\gamma_n][\gamma_{n+1}]^{-1}=[\gamma][\gamma]^{-1}=1 for each n\geq 1.

Now, we use a “telescoping product” to prove that g_{\#}([\ell_{\infty}])\neq 1.

We have

g_{\#}([\ell_{\infty}])=[g\circ\ell_{\infty}]=\left[\prod_{n=1}^{\infty}(\gamma_{n}\cdot\gamma_{n+1}^{-})\right]=[\gamma_1]\left[\prod_{n=2}^{\infty}(\gamma_n\cdot\gamma_{n}^{-})\right]

where the last equality is allowed according to Remark 2. But \left[\prod_{n=2}^{\infty}(\gamma_n\cdot\gamma_{n}^{-})\right]=1 by Lemma 4.

Therefore g_{\#}([\ell_{\infty}])=[\gamma](1)=[\gamma]\neq 1. This completes the proof! \square

Takeaway

There are a few things I hope you can take away from this post. Ultimately, we have taken this important obstruction and teased it apart into different viewpoints. To me, that makes good mathematics.

  1. Because the abelianization of \pi_1(\mathbb{HA},b_0) is isomorphic to \prod_{n=1}^{\infty}\mathbb{Z}/\bigoplus_{n=1}^{\infty}\mathbb{Z} (a highly non-trivial fact), Condition 2. in Theorem 1 looks like a non-abelian generalization of the cotorsion free property defined for abelian groups. In fact, a direct Corollary of 1. \Leftrightarrow 2. is that if \pi_1(X,x_0) is abelian and cotorsion free, then X is homotopically Hausdorff.
  2. Condition 3 in Theorem 1 looks like a closure property – something like subgroup-closure under infinite products…we make this more precise and apply the idea widely in the paper I shamelessly plug at the top of the post.

 

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This entry was posted in Algebraic Topology, Fundamental group, Group homomorphisms, harmonic archipelago, Hawaiian earring and tagged , , . Bookmark the permalink.

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