TITLE:  $\pi_4$ of simply-connected 4-manifold  
QUESTION [15 upvotes]: In Baues "The homotopy category of simply conected 4-manifolds" there is some 
algebraic description of $\pi_4(M^4)$ where $M^4$ is simply-connected closed 4-manifold, but this description is pure algebraic.
And with it there is geometric description of $\pi_3(M^4)$:
$M^4 = \vee S^2 \cup e^4$, and attaching map of $e^4$ is element of $\pi_3(\vee S^2)$. So, $\pi_3(M^4)$ 
is a $\pi_3(\vee S^2)$ with additional relation.
Each element in $\pi_3(\vee S^2)$ has description in terms of linking number of point preimages (circles in $S^3$) of map $S^3\to \vee S^2$.
Does anyone knows geometric description (I think in terms of Whitehead product and typical preimages) of $\pi_4(M^4)$?

REPLY [4 votes]: I don't know exactly how to answer your question, but one observation is that if $M^4 \ncong S^4$, then for any map $\alpha : S^4\to M^4$, $H^4(\alpha):H^4(M^4)\to H^4(S^4)$ is the zero map. This follows because $[M]=a\cup b$ for $a,b\in H^2(M^4)$, by the Poincare conjecture and Poincare duality. Thus, one sees that the Hurewicz map is trivial.
If you have a geometric description of $M^4= \vee S^2 \cup e^4$, then for any point $x\in e^4$, the map $\alpha$ may be homotoped to miss $x$. To see this, make $\alpha$ transverse to $x$. Then the preimage of $x$ is finitely many points, whose orientations cancel since $H_4(\alpha)=0$. Choose arcs in $S^4$ pairing up the points in opposite signs, then since $M$ is simply connected, we may homotope neighborhoods of these arcs rel endpoints to be constant by homotopy extension, and then homotope off of $x$ (this is a version of the Whitney trick). Now by cellular approximation, the map may be homotoped onto $\vee S^2$. So the homotopy groups come from the homotopy groups of the 2-skeletion $\vee S^2$. I'm not certain though how to figure out what relations come from adding the 4-cell $e^4$ though.