# $S^1 \times S^2$ vs $S^1 \vee S^2 \vee S^3$

This is a multi-part problem. Let $X = S^1 \times S^2$ and $Y = S^1 ­\vee S^2 \vee S^3.$

1. Compute $\pi_1$ of those spaces.
2. Do there exist $\phi:S^3 \to X$ and $\psi:X \to S^3$ such that $\psi \phi \simeq 1_{S^3}?$ Hint: use covering space theory and $H_*(S^k;\mathbb{Z}/2).$
3. Using 2. say whether $X$ and $Y$ have the same homotopy type or not.

What I tried:

1. This is easy enough. Using the fact that $\pi_1$ preserves products, we get that $\pi_1(X) \cong \mathbb{Z}.$ Also by Seifert-Van Kampen we see that for a wedge of nice spaces such as these $\pi_1(A \vee B) \cong \pi_1(A) * \pi_1(B),$ so $\pi_1(Y) \cong \pi_1(S^1)*\pi_1(S^2)*\pi_1(S^3) \cong \mathbb{Z}$ also.

2. I'm not too sure about how to proceed here. First I computed $H_i(X) \cong H_i(Y) \cong \mathbb{Z}$ for $i = 0,\ldots,3$ and $0$ otherwise using Künneth for $X$ and the standard M-V sequence argument for $Y.$ If the question asked whether $\phi \psi \simeq 1_{X}$ were possible instead then I'd say no for then the identity map of $H_2(X)$ would factor through $H_2(S^3) = 0.$ As it stands though I'm not sure how to see a contradiction (or how to use covering space theory here...).

3. I haven't really thought about this yet. Maybe we should assume $X \simeq Y$ and somehow conclude that $X \simeq S^3$ to contradict 2.?

• 3) follows from the negative answer to 2). For 2), because $S^3$ is simply connected, you can factor any map $S^3 \to X$ through the universal cover $\tilde X \to X$. – user98602 May 26 '15 at 20:34
• @MikeMiller That is true, but how does it help? It tells us that the map induced from the identity $S^3 \to S^3$ at the level of $\pi_1$ or $H_1$ factors through the zero map, but we already know that that map is zero. – anon May 26 '15 at 20:45
• The identity map on $S^3$ probably shouldn't induce zero on $H_3$. – user98602 May 26 '15 at 20:46
• @MikeMiller Hm, then I think the problem is I don't see why $H_3(\tilde{X}) = 0.$ Do we need explicit knowledge of that space or does it follow from more general considerations? – anon May 26 '15 at 20:48
• It follows from more general considerations (a noncompact $n$-manifold has $H_n(X) = 0$, and because $\pi_1(X)$ is infinite, $\tilde X$ must be noncompact), but the fact that $X = S^2 \times S^1$ makes it much easier to write down the universal cover of $X$. – user98602 May 26 '15 at 20:50

Because $S^3$ is connected, any map $\phi: S^3 \to X$ factors through the universal cover $\tilde X \to X$. Because $\tilde X \cong S^2 \times \mathbb R \simeq S^2$, $H_3(\tilde X) = 0$. So any map $\phi: S^3 \to X$ induces the zero map on $H_3$. In particular, if you picked $\psi: X \to S^3$, then $\psi \phi$ induces zero on $H_3$, and cannot be homotopic to the identity map. So no, there are no such maps.
As in the discussion in the comments, if $\pi_1(X)$ is infinite, $\tilde X$ must be noncompact. Because every noncompact 3-manifold has $H_3(M) = 0$, we see by the same argument that if $X$ is any 3-manifold with infinite fundamental group, there is not a pair of maps $S^3 \to X \to S^3$ whose composition is homotopic to the identity. Indeed I would bet this is true for any 3-manifold $X$ that's not $S^3$, but I don't see an obvious proof.
3) now follows from the negative solution to 2), because $S^1 \vee S^2 \vee S^3$ does support such a map.
• Oh! Then for 3., the composition $S^3 \to S^1 \vee S^2 \vee S^3 \to S^3$ where the first map is inclusion and the second map sends $S^1$ and $S^2$ to the basepoint on $S^3$ and restricts to the identity on $S^3$ works. – anon May 26 '15 at 21:01