Let $W(t)$ be a Brownian Motion for $t\geq 0$, and $\tau_m$ be the first passing time of $W(t)$ above $m$ such that $\tau_m=\text{inf}\{t: B_t\geq m\}$. Then, according to Shreve's Stochastic Calculus for Finance II equation (3.7.1):
$$\mathbb{P}(\tau_m \leq t, W(t)\leq w)=\mathbb{P}(W(t)\geq 2m-w)$$
given $w\leq m$ and $m>0$. The book motivates the equality by taking all Brownian paths that satisfy $\tau_m \leq t, W(t)\leq w$ and inverting them after they cross $m$ at $\tau_m$, which would then satisfy $\tau_m \leq t, W(t)\geq 2m-w$. From what I understand, this argument creates a bijection between the paths included in each probability statement above.
However, what I am struggling to understand is why couldn't we make the same bijective argument for any $r\in\mathbb{R}$ instead of $2m-w$. For example, what prevents us from stating that $$\mathbb{P}(\tau_m \leq t, W(t)\leq w)=\mathbb{P}(W(t)\geq 3m-w)$$ I know this is a bit of a stretch, but my reasoning is that there are uncountably infinite Brownian paths satisfying $\tau_m \leq t, W(t)\leq w$ and likewise satisfying $W(t)\geq 3m-w$. What prevents us from coming up with some bijection between these two set of paths?
