 Solution

Use the standard sl2 Lie algebra relations $[h,e]=2e$, $[h,f]=-2f$ and $[e,f]=h$. The classification of finite-dimensional sl2 representations says that every finite-dimensional complex module is a direct sum of $L(m)$, $m\in {\mathbb{Z}}_{\ge 0}$, where $L(m)$ has weights $m,m-2,\dots ,-m$. The raising operator $e$ and lowering operator $f$ move weights by two and minus two. On $L(m)$ each therefore has $(m+1)$st power zero. There is consequently one integer $N$ such that $e^N=f^N=0$ on all of $V$.

The matrix exponentials are thus finite polynomials in nilpotent operators:Each is a well-defined endomorphism, and its inverse is obtained by negating the argument. Their product is therefore a linear automorphism,This Weyl reflection lift in an sl2 representation is algebraically defined by finite sums; no choice of topology or convergence of an infinite series is needed.