# Why is a simply connected homology sphere a topological sphere?

I post this for a friend who currently doesn’t have access to this site.

It is about an implication in the last paragraph of the following paper:

The question is about the last sentence:

Since $$\tilde{M}$$ is simply connected, it is also a topological sphere.

Why is this true? Is it easy to work out a proof of this by hand ? If not may I ask a reference from which the question simply follows?

Sorry if it is a well-known fact in algebraic topology (if so may you point out a reference?)

Thanks a lot !

• I have not looked at the paper, but this may be the key point: en.wikipedia.org/wiki/Generalized_Poincar%C3%A9_conjecture Nov 25 '21 at 9:30
• It is more or less topological Poincare conjecture Nov 25 '21 at 18:01
• I am curious why someone doesn't have access to this site. Even Valery Fabrikant visits. (Привет!) Nov 28 '21 at 10:03
• @BenMcKay: maybe en.wikipedia.org/wiki/Great_Firewall Nov 28 '21 at 14:36

There are few steps.

The first one. A simply connected homology sphere is a homotopy sphere actually. It follows from the combination of the Whitehead and Hurewicz theorems. By the Hurewicz theorem, $$\pi_n(X) \cong H_n(X) \cong \mathbb Z$$. Therefore, there is a map inducing homology isomorphism. And by the Whitehead theorem it is a homotopy equivalence.

The second one. Any homotopy sphere is actually topological sphere. It follows from classification theorem in dimension 2. It follows from Poincare conjecture in dimension 3. And finally, in dimension n $$\geqslant 4$$ it follows from the generalized Poincaré conjecture, proved by Smale 1960-61 for n ≥ 5 and ca. 1982 by Michael Freedman for n = 4.

• Perhaps it's worth noting that the result that "any homotopy sphere is actually a topological sphere" is often called "the generalised Poincaré conjecture" (en.wikipedia.org/wiki/Generalized_Poincaré_conjecture), and its proof in dimension 4 is quite different from the proof in dimensions $\geq 5$.
– HJRW
Nov 28 '21 at 16:49
• @HJRW I thought that the only difference in dimension 4 is an application of the Whitney trick. Nov 30 '21 at 17:54
• No: see this recent Quanta article to get an idea of how much more difficult the 4d case is. quantamagazine.org/…
– HJRW
Nov 30 '21 at 20:50