## Homotopically Hausdorff Spaces (Part II)

In my post homotopically Hausdorff spaces (Part I), 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. To Appear in the Journal of Topology and Analysis. 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.

### (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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