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

Question

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

Compute $\pi_1$ of those spaces.
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 $H_*(S^k;\mathbb{Z}/2).$
Using 2. say whether $X$ and $Y$ have the same homotopy type or not.

What I tried:

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.
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...).
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.?

@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 $H_3$. — , 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 $X$. — , May 26 '15 at 20:50
@MikeMiller In this case I think $\tilde{X} = \mathbb{R} \times S^2,$ and indeed this implies that $H_3(\tilde{X}) = 0.$ — anon, May 26 '15 at 20:55

score 4 · Accepted Answer · answered May 26 '15 at 20:56

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
If $S^3 \to X \to S^3$ is homotopic to the identity, then $H_3(X; \mathbb{Z}) \cong \mathbb{Z}$ so $X$ is closed and orientable. Moreover, the map $S^3 \to X$ is degree one (with the appropriate choice of orientations), so $\pi_1(S^3)$ surjects onto $\pi_1(X)$. As $S^3$ is simply connected, it follows that $X$ is simply connected and hence $X = S^3$ by the resolution of the Poincaré conjecture. — Michael Albanese, Feb 09 '21 at 12:08

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

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