In recent posts, I’ve been writing about the behavior of fundamental groups of the most fundamental “wild” spaces. We did a decent amount of work in a two-part post to convince ourselves that the fundamental group of the Hawaiian earring is not free (Part I and Part II). What we have going for us is that we actually have an explicit combinatorial description of the Hawaiian earring group. This allows us to make arguments without having to discuss loops (this is a big-time theme in algebraic topology).

Now I’m going to work through an interesting property of another important group: the fundamental group of the harmonic archipelago

In particular, I’m going to try to convince you that **every homomorphism** **to the additive group of integers is trivial**. In otherwords . This is a bit strange considering is uncountable and torsion free.

You can construct the harmonic archipelago by drawing a copy of the Hawaiian earring on a solid disk and pushing up a hill of height one in between each of the hoops. Notice that the Hawaiian earring is naturally a subspace of . So is basically just a really bumpy disk that differs topologically from a disk only at the one wild point of the Hawaiian earring.

From the inclusion , we have an induced homomorphism . In my first post on the harmonic archipelago, I argued that is surjective and hence that is a natural quotient of . In particular, if is the homotopy class of the loop going once around the n-th circle of the Hawaiian earring, then is the conjugate closure of the set . This is essentially because is homotopy equivalent to the space obtained by attached 2-cells to using the loops , as attaching maps. Unfortunately, the characterization as is not helpful for understanding the combinatorial structure of the group so I discussed a more practical characterization as a direct limit of Hawaiian earring groups at the end of this post.

**Homomorphisms out of **

Let’s first recall the two key facts that we needed to show is not a free group. These facts are Lemma 4 from Part I and Lemma 6 from Part II respectively.

**Fact I:** For every homomorphisms , there is an such that for all .

**Fact II:** If is a homomorphism such that for all , then

**Theorem 1:** Every homomorphism is trivial.

*Proof.* Consider the composition . According to Fact I, there is an such that for all . But since for all , we have for all . Now according to Fact II, we have . Since is surjective, .

**Corollary 2:** is not isomorphic to a free group.

*Proof.* There is no non-trivial free group such that every homomorphism is trivial. Therefore is not free.

One might be tempted to conjecture that the groups and are isomorphic since they are both uncountable, torsion free, and we get by introducing countably many relations in the uncountable group . Now it is clear that this cannot be the case.

**Corollary 3:** is not isomorphic to .

*Proof.* Non-trivial homomorphisms exist. In fact, as we saw in my last post. For example, the retraction to the unit circle which collapses all but the first circle of induces a surjective homomorphism .

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