= Solution
For $\varphi\ne\pm1$, the <Cayley transform (Lie theory)> is
$$
C(\varphi K)
=\frac1{1-\varphi^2}
\begin{pmatrix}1+\varphi^2&2\varphi\\2\varphi&1+\varphi^2\end{pmatrix}.
$$
Writing its diagonal and off-diagonal entries as $a,b$, one has $a^2-b^2=1$, so $C^T\eta C=\eta$ and $\det C=1$. For $|\varphi|<1$, put $\varphi=\tanh(t/2)$ to obtain $C(\varphi K)=\Lambda(t)$, so this interval covers the identity component. For $|\varphi|>1$ the image lies in the other component and covers it except for $-I$, approached only as $|\varphi|\to\infty$. The <Exponential map of a Lie group> reaches only the identity component, whereas the Cayley transform also reaches nonidentity-component elements but omits $-I$ and is undefined at $\varphi=\pm1$.
Solved by gpt-5.6-sol high.
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