Let $B_t$ be a $1$-dimensional Brownian motion. By time inversion, $(B_{t-u}-B_t)_{u\in [0,t]}$ is again a BM, and we get $$(B_t, \sup_{0\le u\le t}(B_u - B_t)) \sim (-B_t, \sup_{0\le u\le t}B_u)).$$
This is part of a solution to a problem from Schilling's Brownian Motion. I don't understand how the two have the same distributions. I know that $B_t \sim -B_t$, and it seems like $\sup_{0\le u\le t}(B_u - B_t)) \sim \sup_{0\le u\le t}B_u$(which I'm not sure exactly why). But even in this case, how do we guarantee that the joint distributions are the same, since the components are not independent here. I would greatly appreciate an explanation.
Define $B_u' = -(B_t - B_{t-u})$ then we know that $(B_u')_{0 \leq u \leq t}$ is a Brownian motion. We also have that $B_t = -B_t'$ and $$\sup_{0 \leq u \leq t}(B_u - B_t) = \sup_{0 \leq u \leq t}(-(B_t - B_{t-u})) = \sup_{0 \leq u \leq t} B_u'.$$
So $$(B_t, \sup_{0 \leq u \leq t} (B_t,\sup_{0 \leq u \leq t}(B_u - B_t)) = (-B_t', \sup_{0 \leq u \leq t} B_u')$$
thereby completing the proof.