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$\DeclareMathOperator\id{id}$Let $\Omega^2(S^2,x)$ be the set of basepoint-preserving maps $S^2\to S^2$ (with basepoint $x=(1,0,0)$). Instead of taking the constant map $S^2\to S^2$ as the basepoint of $\Omega^2(S^2)$, let's take the identity map $\id_{S^2}$ to be the basepoint.

Certainly, it is the case that $\pi_3(S^2,x)\cong \pi_1(\Omega^2(S^2),\id_{S^2})$ since loop spaces have homotopy equivalent path components. Also, if $X$ is the path component of $\id_{S^2}$ in $\Omega^2(S^2)$, then $\pi_1(\Omega^2(S^2),\id_{S^2})\cong H_1(X)$ by the Hurewicz map. Hence, $\pi_1(\Omega^2(S^2),\id_{S^2})=H_1(X)$ is infinite cyclic.

Let $f:S^1\to \Omega^2(S^2)$ be the loop based at $\id_{S^2}$, which rotates $S^2$ once around the $x$-axis. This map is a low-dimensional case of the $J$-homomorphism and its homotopy class apparently generates $\pi_1(\Omega^2(S^2),\id_{S^2})$.

The map $f$ itself is very simple and highly geometric but the connection to the Hopf map gets a bit muddled. What is a direct, "elementary" proof of why the homotopy class of $f$ generates $\pi_1(\Omega^2(S^2),\id_{S^2})$? By "elementary," I mean some argument using elementary tools from homology theory, geometry, or homotopy theory that perhaps connects back to the Hopf map (or not) and could be understood by someone who does not know general results characterizing the image of the $J$-homomorphism.

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    $\begingroup$ In this case, if you already know $\pi_3(S^2) =\mathbb{Z}$ to characterize the image of the J-homomorphism really just requires knowledge of the Euler class of a vector bundle and of a spherical fibration. These are pretty easy to define. $\endgroup$
    – Connor Malin
    Commented May 27, 2021 at 16:17
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    $\begingroup$ This is not too hard to see from the point of view of the Pontryagin-Thom construction. You adjoin the map into a map $S^3 \to S^2$, and then look at the inverse image of a generic point (not the base point or the fixed points of the rotation). You also keep track of the framing from $\mathbb{R}^2$ at that point. The inverse image is a framed unknot whose framing "twists" around once, and so has self linking number (i.e. Hopf invariant) equal to one. $\endgroup$
    – Chris Schommer-Pries
    Commented May 28, 2021 at 14:59
  • $\begingroup$ The argument I had in mind only worked for the stabilization, i.e. $\pi_1(\Omega^3 S^3)$. In that case, it is just factorizing the second Stiefel-Whitney class $[S^2,BSO(2)] \rightarrow [S^2,B \Omega^3 S^3] \rightarrow H^2(S^2,\mathbb{Z}/2)$ and noticing that this is isomorphic to a factorization $\mathbb{Z} \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/2$ where the entire composition is surjective. $\endgroup$
    – Connor Malin
    Commented May 29, 2021 at 21:29

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