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redid the diagram. added reference
Vitali Kapovitch
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Let me try to summarize various observations from the comments and put it all together.

Let $d\ge 5$ and let $f: S^4\to BSO(d)$ be a generator of $\pi_4(BSO(d))\cong \mathbb Z$.

Let $\gamma^d$ be the universal orientable $\mathbb R^d$ vector bundle over $BSO(d)$.

Then we have the following diagram extending the one by the OP. It is easily seen to commute by naturality.

enter image description here

Note that as indicated on the diagram the bottom $\alpha$ and the right $f^*$ are isomormphisms.

Observe that $f^*(p_1(\gamma^d))=p_1(f^*(\gamma^d))$. It is well known that $p_1$ of any vector bundle over $S^4$ is divisible by 2 (if we identify $H^4(\mathbb S^4)$ with $\mathbb Z$) and in fact every even number is realized. In particular $p_1$ of the Hopf $\mathbb R^4$ bundle is $\pm 2$. The above is done in detail for example in Milnor's classical paper on exotic $7$-spheres.

Since the two arrows ending in $Hom(\pi_4(\mathbb S^4), \mathbb Z)$ are isomorphisms this implies that the image of $\alpha$ is exactly $2\mathbb Z$ (we again identify $Hom(\pi_4(BSO(d)),\mathbb Z)$ with $\mathbb Z$).

Moreover this also shows that $\alpha(p_1)=\pm 2\beta(x)$ (recall that $x\in Hom(\pi_3(SO(d),\mathbb Z)$ is a generator).

Furthermore this shows that the clutching map of the stabilized Hopf bundle is exactly the generator of $\pi_3(SO(d))$ since if it wasn't then $p_1$ of the Hopf bundle would be a nontrivial multiple of $2$.

The clutching map of the Hopf bundle is just the identity map $\mathbb S^3=Sp(1)\to Sp(1)\cong SU(2)\subset SO(4)$ and therefore the generator of $\pi_3(SO(d))$ comes from the inclusion $SU(2)\to SO(4)\to SO(d)$.

This can also be seen in other ways without using Pontryagin classes. For example, recall the following well known picture of $SO(4)$.

Think of $\mathbb R^4$ as quaternions and $\mathbb S^3$ as $Sp(1)$ then consider the ineffective isometric linear $\mathbb S^3\times \mathbb S^3$ action on $\mathbb R^4$ given by $(q_1,q_2)(v)=q_1vq_2^{-1}$. The kernel is $\mathbb Z_2=\{\pm (1,1)\}$ and we get $SO(4)=(\mathbb S^3\times \mathbb S^3)/\mathbb Z_2$. The first $\mathbb S^3$ commutes with multiplication by $i$ on the right and the second with multiplication by $i$ on the left. This identifies the two $Sp(1)$'s with two different embeddings of $SU(2)$ in $SO(4)$. The right $Sp(1)$ is easily seen to correspond to the standard embedding. The two $SU(2)$' give generators of $\pi_3(SO(4))\cong\mathbb Z\oplus \mathbb Z$. Now look at the inclusion $SO(4)\to SO(5)$ and the induced map on $\pi_3$. From the homotopy sequence of the bundle $SO(4)\to SO(5)\to S^4$ we get that the map $\pi_3(SO(4))\to \pi_3(SO(5))\cong \mathbb Z$ is onto.

It's easy to see by looking at the Weyl groups that the two $SU(2)$'s become conjugate in $SO(5)$. Hence they have the same image in $\pi_3$ which must be the generator else $\pi_3(SO(4))\to \pi_3(SO(5))$ would not be onto.

Lastly, this could be seen using general theory by computing the Dynkin index of $SU(2)$ in $SO(5)$. Given a simple compact Lie group $G$ and its simple subgroup $H$ normalize their Killing forms $(\cdot, \cdot)_G$ and $(\cdot, \cdot)_H$ so that the longest roots have length $\sqrt 2$. Then the restriction of $(\cdot, \cdot)_G$ to the Lie algebra of $H$ is proportional to $(\cdot, \cdot)_H$. The coefficient is called the Dynkin index of $H$ in $G$. It is always an integer and up to sign is equal to the coefficient in the induced map $\pi_3(H)\cong \mathbb Z\to \pi_3(G)\cong \mathbb Z$. For the proof see the book "Topology of transitive transformation groups" by Onishchik.

It is not hard to compute that the Dynkin index of $SU(2)$ in $SO(5)$ is 1 but won't do it as we have two other proofs already.

Vitali Kapovitch
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