A manifold with $\pi_1 = \pi_2 = \pi_4 = 0, \pi_3 = \mathbb{Z}$

I am looking for an example of a closed connected smooth manifold $M$ with $\pi_1(M) = \pi_2(M) = \pi_4(M) = 0$ and $\pi_3(M) = \mathbb{Z}$, if such a thing exists.

My motivation is to represent the third integral cohomology of a 4-manifold $X$ by (homotopy classes of) maps to some manifold. Since $S^1$ is a $K(\mathbb{Z}, 1)$ and $\mathbb{CP}^\infty$ is a $K(\mathbb{Z}, 2)$, we have $H^1(X ; \mathbb{Z}) = [X, K(\mathbb{Z}, 1)] = [X, S^1]$ and, by cellular approximation, $H^2(X;\mathbb{Z}) = [X, \mathbb{CP}^\infty] = [X, \mathbb{CP}^2]$. So we can represent first and second cohomology of $X$ by maps into manifolds; my question just asks about doing so for the next higher degree.

Indeed, if we had a manifold $M$ satisfying $\pi_1(M) = \pi_2(M) = \pi_4(M) = 0$, $\pi_3(M) = \mathbb{Z}$, then we could form a $K(\mathbb{Z}, 3)$ from $M$ by attaching 6-cells and higher to kill off $\pi_5$ and higher. So, we would have that the five-skeleta of these two spaces coincide, $M^{(5)} = K(\mathbb{Z}, 3)^{(5)}$, and thus by cellular approximation $$[X, M] = [X, M^{(5)}] = [X, K(\mathbb{Z}, 3)^{(5)}] = [X, K(\mathbb{Z}, 3)] = H^3(X;\mathbb{Z}).$$

If such an $M$ exists, its dimension must be at least 8. Indeed, first observe that $M$ is orientable, since it is simply connected.

All orientable surfaces have trivial $\pi_3$, except for the sphere which has non-trivial $\pi_2$, so that rules out dimension 2.

Any 2-connected 3-manifold is a homotopy sphere, so this rules out dimension 3 since $\pi_4(S^3) = \mathbb{Z}_2$.

Hurewicz implies that $H_1(M;\mathbb{Z}) = H_2(M;\mathbb{Z}) = 0$ and $H_3(M;\mathbb{Z}) = \mathbb{Z}$. Poincaré duality now shows that $M$ cannot have dimension 4 or 5.

Since the Euler characteristic of an oriented $4k+2$ dimensional manifold is even, we can rule out dimension 6 by applying Poincaré duality and computing the Euler characteristic of such an $M$ to be 1.

Hurewicz also tells us that $\pi_4(M)$ surjects onto $H_4(M;\mathbb{Z})$, so $H_4(M;\mathbb{Z}) = 0$. This rules out $M$ being a 7-manifold because of Poincaré duality again.

There is actually an $8$ dimensional manifold which has your desired homotopy groups: The Lie group $SU(3)$.

In fact, the result follows from Bott Periodicty. More specifically, for $U = \bigcup_n U(n)$, the union of the unitary groups, Bott periodicity shows that $\pi_k(U)$ only depends on the parity of $k$: when $k$ is even, this group is isormphic to $\mathbb{Z}$, and when $k$ is odd it is trivial.

But from the inclusion $U(3)\subseteq U$ is at least 5 connected, so $\pi_k(U(3))\cong \pi_k(U)$ for $k\leq 4$. Finally, since $SU(3)\times S^1\cong U(3)$, $SU(3)$ must have the right homotopy groups.

Consider the $n$-ball for $n$ large. Attach a 3-handle and call the resulting manifold $M$. It should be relatively easy to see the map $\pi_4(\partial M) \to \pi_4(M)$ is surjective (one way is by turning the cobordism "upside down" and thinking about the 3-handle as an n-3 handle and the 0-handle as an n-handle). Now attach 5-handles to the boundary along all the generators of $\pi_4(\partial M)$ (which can be represented by smoothly framed embeddings by Whitney when $n$ is large enough).

Now, the resulting manifold with boundary has a handle decomposition (Morse function) with one 0-handle, one 3-handle and one 5-handle (its easy to see the $\pi_4$ you are killing is actually $\Bbb Z/2$), so its double has one 0-handle, one 3-handle, one 5-handle, one $n$-5 handle, one $n$-3 handle and one $n$ handle. For $n$ large enough, only the first three handles will affect the desired homotopy groups.

This is a typical argument to construct compact smooth manifolds with any $k$-type you want. For your case the smallest $n$ where this argument works is $10$.