For a general stopping time $\tau$, we define $\hat{B}(t)=B(t)1_{\{t\leq \tau\}}+(2B(\tau)-B(t))1_{\{t>\tau\}}$.
I know the strong Markov property $B'(t)=(t+\tau)-B(\tau)$ is also a Brownian motion and it's independent of $F_{\tau}$. And I also know to get $\hat{B}$, it involves gluing of $B$ and $B'$. The main goal is to prove $\hat{B}$ is also a Brownian motion.I tried proving independence and stationary by splitting the stopping time, and finite dimensional distribution method, but both didn't work.
We follow Theorem 5.2.13 in Probability theory II, 6.13 Theorem in R.Schilling and Proving the reflection principle of Brownian motion.
On the event that $\tau$ is finite, by the strong Markov property the path
$$(g_{1}(t)=B_{t+\tau}-B_{\tau})_{t\geq 0}, (g_{2}(t)=B_{\tau}-B_{t+\tau})_{t\geq 0}$$
is a Brownian motion independent of $(B_{t})_{t\leq \tau}$. We define the concatenation mapping
\begin{equation} \Phi : \mathbf{C}_{(0)}\times[0,\infty)\times\mathbf{C}_{(0)} \rightarrow \mathbf{C}_{(0)}, \end{equation}
where $\mathbf{C}_{(0)} := \{f \in \mathbf{C}[0,\infty) : f(0) = 0\}$, thus
\begin{equation}\Phi(f,t,g) := \begin{cases} f(s), & 0 \leq s < t \\ f(t) + g(s - t), & t \leq s < \infty \end{cases} \end{equation}
This map $\Phi$ is continuous if we equip $\mathbf{C}_{(0)}\times[0,\infty)\times\mathbf{C}_{(0)}$ with the topology of locally uniform convergence (in $\mathbf{C}_{(0)}$) and pointwise convergence (in $[0,\infty)$). Therefore, it is $\mathcal{B}(\mathbf{C}_{(0)})\otimes\mathcal{B}[0,\infty)\otimes\mathcal{B}(\mathbf{C}_{(0)})/\mathcal{B}(\mathbf{C}_{(0)})$-measurable.
This continuity is easy to see by taking sequences $(f_{n},t_{k},g_{m})\rightrightarrows (f,t,g)$ and studying $\Phi(f_{n},t_{k},g_{m}), \Phi(f,t,g)$ by splitting the supremum over $t_{k}$.
From here the result follows easily by using the $\Phi$-map to concatenate $g_{1},g_{2}$ separately to $(B(t))_{t\leq \tau}$.