[proofplan]
We compute the commutator by applying it to an arbitrary vector $v\in\mathbb R^3$. The defining property of the hat map converts matrix multiplication into repeated cross products. A direct coordinate verification of the vector triple product identity then shows that the commutator action is exactly the action of $\widehat{a\times b}$. Since linear maps on $\mathbb R^3$ are equal when they agree on every vector, this proves the bracket identity; the [Lie algebra](/page/Lie%20Algebra) isomorphism follows from the linear bijectivity of the hat map.
[/proofplan]
[step:Convert the matrix commutator into a triple cross product]
Let $a,b\in\mathbb R^3$ be arbitrary, and let $v\in\mathbb R^3$ be arbitrary. The bracket on $\mathfrak{so}(3)$ is the commutator bracket, so
\begin{align*}
[\widehat a,\widehat b]v=(\widehat a\widehat b-\widehat b\widehat a)v.
\end{align*}
Using the defining property $\widehat x\,w=x\times w$ of the hat map, first with $x=b$ and then with $x=a$, gives
\begin{align*}
\widehat a\widehat b v=a\times(b\times v).
\end{align*}
Similarly,
\begin{align*}
\widehat b\widehat a v=b\times(a\times v).
\end{align*}
Therefore
\begin{align*}
[\widehat a,\widehat b]v=a\times(b\times v)-b\times(a\times v).
\end{align*}
[/step]
[step:Use the vector triple product identity to identify the resulting action]
We use the vector triple product identity
\begin{align*}
x\times(y\times z)=y(x\cdot z)-z(x\cdot y)
\end{align*}
for $x,y,z\in\mathbb R^3$. Applying it twice gives
\begin{align*}
a\times(b\times v)=b(a\cdot v)-v(a\cdot b).
\end{align*}
Also,
\begin{align*}
b\times(a\times v)=a(b\cdot v)-v(a\cdot b).
\end{align*}
Subtracting these two identities yields
\begin{align*}
a\times(b\times v)-b\times(a\times v)=b(a\cdot v)-a(b\cdot v).
\end{align*}
The same vector triple product identity applied to $(a\times b)\times v$ gives
\begin{align*}
(a\times b)\times v=b(a\cdot v)-a(b\cdot v).
\end{align*}
Hence
\begin{align*}
[\widehat a,\widehat b]v=(a\times b)\times v.
\end{align*}
[guided]
The goal of this step is to recognize the expression obtained from the commutator as another hat-map action. Since $\widehat c$ acts on a vector $v$ by $c\times v$, we want to rewrite
\begin{align*}
a\times(b\times v)-b\times(a\times v)
\end{align*}
in the form $c\times v$ for a specific vector $c$.
We use the vector triple product identity: for all $x,y,z\in\mathbb R^3$,
\begin{align*}
x\times(y\times z)=y(x\cdot z)-z(x\cdot y).
\end{align*}
With $x=a$, $y=b$, and $z=v$, this gives
\begin{align*}
a\times(b\times v)=b(a\cdot v)-v(a\cdot b).
\end{align*}
With $x=b$, $y=a$, and $z=v$, it gives
\begin{align*}
b\times(a\times v)=a(b\cdot v)-v(b\cdot a).
\end{align*}
Since the Euclidean dot product is symmetric, $b\cdot a=a\cdot b$. Therefore, subtracting the two displayed identities cancels the two terms involving $v(a\cdot b)$ and leaves
\begin{align*}
a\times(b\times v)-b\times(a\times v)=b(a\cdot v)-a(b\cdot v).
\end{align*}
Now apply the same triple product identity to $(a\times b)\times v$. This is the case $x=a\times b$, $y$ not used directly is better handled by the equivalent identity
\begin{align*}
(x\times y)\times z=y(x\cdot z)-x(y\cdot z).
\end{align*}
Taking $x=a$, $y=b$, and $z=v$, we obtain
\begin{align*}
(a\times b)\times v=b(a\cdot v)-a(b\cdot v).
\end{align*}
Combining this with the previous computation gives
\begin{align*}
a\times(b\times v)-b\times(a\times v)=(a\times b)\times v.
\end{align*}
Using the commutator computation from the preceding step, we conclude
\begin{align*}
[\widehat a,\widehat b]v=(a\times b)\times v.
\end{align*}
[/guided]
[/step]
[step:Conclude equality of the matrices from equality on all vectors]
By the defining property of the hat map applied to the vector $a\times b$, we have
\begin{align*}
\widehat{a\times b}\,v=(a\times b)\times v.
\end{align*}
The previous step proves that
\begin{align*}
[\widehat a,\widehat b]v=\widehat{a\times b}\,v
\end{align*}
for every $v\in\mathbb R^3$. Since two linear maps $\mathbb R^3\to\mathbb R^3$ that agree on every vector are equal, it follows that
\begin{align*}
[\widehat a,\widehat b]=\widehat{a\times b}.
\end{align*}
[/step]
[step:Identify the hat map as the Lie algebra isomorphism]
The hat map $H:\mathbb R^3\to\mathfrak{so}(3)$, $a\mapsto\widehat a$, is linear and bijective by its standard coordinate description. The bracket identity just proved says precisely that for all $a,b\in\mathbb R^3$,
\begin{align*}
H(a\times b)=[H(a),H(b)].
\end{align*}
Thus $H$ is a linear bijection preserving brackets. Therefore $H$ is an isomorphism from the Lie algebra $(\mathbb R^3,\times)$ to the Lie algebra $\mathfrak{so}(3)$.
[/step]
