[proofplan] The strategy is to identify the same algebraic object as the unique solution of two different controlled differential equations and read off the desired identity. Set $A := S(x)_{[a, b]}$ and consider the CDE driven by the time-reversed path $\overleftarrow{x}_t := x_{a + b - t}$, with initial datum $A$. Its unique solution is $U_t = A \cdot S(\overleftarrow{x})_{[a, t]}$. Now run time backwards by setting $V_t := U_{a + b - t}$; a direct computation (using $d\overleftarrow{x}_{a + b - t} = -dx_t$ in the Young sense) shows that $V$ solves the CDE driven by $x$ with terminal condition $V_b = A$. Uniqueness of the CDE driven by $x$ — applied with the boundary value $V_b = S(x)_{[a, b]}$ — forces $V_t = S(x)_{[a, t]}$, hence $V_a = \mathbf{1}$. Translating back, $V_a = U_b = A \cdot S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]} = \mathbf{1}$. The other identity follows from $\overleftarrow{\overleftarrow{x}} = x$. [/proofplan]

[step:Set up the CDE driven by $\overleftarrow{x}$ with initial datum $A := S(x)_{[a, b]}$] Define the time-reversed path \begin{align*} \overleftarrow{x}: [a, b] &\to V \\ t &\mapsto x_{a + b - t}. \end{align*} Since $t \mapsto a + b - t$ is a $C^\infty$ bijection $[a, b] \to [a, b]$, it preserves $p$-variation, so $\overleftarrow{x} \in C_p([a, b], V)$ with $\|\overleftarrow{x}\|_{p\text{-var}; [a, b]} = \|x\|_{p\text{-var}; [a, b]}$. Set $A := S(x)_{[a, b]} \in T((V))$ and let $E$ be a Banach subalgebra of $T((V))$ containing $\mathcal{S}_p \cup \{A\}$ — such as $T(V)$ itself with a suitable norm, or the closed subalgebra generated by signatures. By the [Signature as Solution of a CDE](/theorems/2497) applied to the path $\overleftarrow{x}$ with initial datum $A$, the CDE \begin{align*} dU_t = U_t \cdot d\overleftarrow{x}_t \quad \text{on } [a, b], \qquad U_a = A, \end{align*} admits the unique solution \begin{align*} U_t = A \cdot S(\overleftarrow{x})_{[a, t]} \quad \text{for } t \in [a, b]. \end{align*} In particular, \begin{align*} U_a &= A \cdot S(\overleftarrow{x})_{[a, a]} = A \cdot \mathbf{1} = A, \\ U_b &= A \cdot S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]}. \end{align*}[/step]

[guided]We launch the proof from the solution-of-a-CDE characterisation of the signature, which is the most flexible computational handle we have on the signature. The time-reversed path is \begin{align*} \overleftarrow{x}: [a, b] &\to V \\ t &\mapsto x_{a + b - t}, \end{align*} i.e. $\overleftarrow{x}$ traces out the same image as $x$ but starts at $x_b$ and ends at $x_a$. Why does $\overleftarrow{x} \in C_p$? The map $\rho: t \mapsto a + b - t$ is a smooth diffeomorphism of $[a, b]$ (with $\rho \circ \rho = \operatorname{id}$). It preserves $p$-variation: for any partition $a = s_0 < s_1 < \cdots < s_N = b$, \begin{align*} \sum_{i=0}^{N-1} |\overleftarrow{x}_{s_{i+1}} - \overleftarrow{x}_{s_i}|^p = \sum_{i=0}^{N-1} |x_{a + b - s_{i+1}} - x_{a + b - s_i}|^p, \end{align*} which is exactly the $p$-variation sum for $x$ over the reversed partition $a = a + b - s_N < \cdots < a + b - s_0 = b$. Hence $\|\overleftarrow{x}\|_{p\text{-var}} = \|x\|_{p\text{-var}} < \infty$. Define $A := S(x)_{[a, b]}$. We need a Banach algebra $E$ such that $A \in E$, $\mathcal{S}_p \subseteq E$, and $E$ is a subalgebra of $T((V))$. The standard choice is the closure (in some natural norm) of the polynomial algebra over the signatures, or simply $E = T_N(V)$ truncated at a large enough level — for the present argument any such $E$ works since the relevant identities hold level-wise. By the [Signature as Solution of a CDE](/theorems/2497) **applied to the driving path $\overleftarrow{x}$** (which lies in $C_p([a, b], V)$ as just verified) with initial condition $A \in E$, the CDE \begin{align*} dU_t = U_t \cdot d\overleftarrow{x}_t, \qquad U_a = A, \end{align*} has the unique solution in $E$: \begin{align*} U_t = A \cdot S(\overleftarrow{x})_{[a, t]}. \end{align*} The starting value is $U_a = A \cdot S(\overleftarrow{x})_{[a, a]} = A \cdot \mathbf{1} = A$, as required. The key data point we need is the terminal value: \begin{align*} U_b = A \cdot S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]}. \end{align*} This is exactly the left-hand side of the identity we want to prove. Our task is now to show $U_b = \mathbf{1}$.[/guided]

[step:Reverse time to obtain $V_t := U_{a + b - t}$ and identify its CDE]Define \begin{align*} V: [a, b] &\to E \\ t &\mapsto U_{a + b - t}. \end{align*} Then $V_a = U_b$ and $V_b = U_a = A = S(x)_{[a, b]}$. We compute the differential of $V$. By the chain rule for Young integration applied to the substitution $\sigma = a + b - t$ (so $d\sigma = -dt$): \begin{align*} V_t - V_s = U_{a + b - t} - U_{a + b - s} = -\bigl( U_{a + b - s} - U_{a + b - t} \bigr) = -\int_{a + b - t}^{a + b - s} U_\sigma \cdot d\overleftarrow{x}_\sigma, \end{align*} where the second equality holds for $s \le t$ (so $a + b - t \le a + b - s$) and the integral is the Young integral. Substituting $\sigma = a + b - r$ (with $d\sigma = -dr$ and bounds reversing) inside the Young integral, \begin{align*} \int_{a + b - t}^{a + b - s} U_\sigma \cdot d\overleftarrow{x}_\sigma = \int_t^s U_{a + b - r} \cdot d\overleftarrow{x}_{a + b - r} \cdot (-1) = -\int_t^s V_r \cdot dx_r = \int_s^t V_r \cdot dx_r, \end{align*} where we used $\overleftarrow{x}_{a + b - r} = x_{a + b - (a + b - r)} = x_r$, so $d\overleftarrow{x}_{a + b - r} = dx_r$ in the Young sense (the change-of-variables formula for Young integrals applies since the substitution is smooth and monotonically reverses orientation; the resulting sign flip is absorbed by the bound reversal). Therefore \begin{align*} V_t - V_s = -\int_s^t V_r \cdot dx_r \cdot (-1) = -\biggl( -\int_s^t V_r \cdot dx_r \biggr) = \int_s^t V_r \cdot dx_r, \end{align*} so $V$ solves the integral equation \begin{align*} V_t - V_s = \int_s^t V_r \cdot dx_r \qquad \text{for all } a \le s \le t \le b, \end{align*} that is, the CDE \begin{align*} dV_t = V_t \cdot dx_t \quad \text{on } [a, b] \end{align*} with the (terminal, not initial) condition $V_b = A = S(x)_{[a, b]}$.[/step]

[guided]The geometric idea is simple: $U$ runs forward in time on $[a, b]$ driven by $\overleftarrow{x}$, and $\overleftarrow{x}$ is itself $x$ run backwards. So if we run $U$ backwards in time, we should be running $x$ forwards. Let's make this precise. Define \begin{align*} V: [a, b] &\to E \\ t &\mapsto U_{a + b - t}. \end{align*} The endpoints swap: $V_a = U_b$ and $V_b = U_a = A$. To compute how $V$ evolves, we need the **change-of-variable formula for Young integrals**. Let $\rho: [a, b] \to [a, b]$, $\rho(t) = a + b - t$, be the time-reversal map. Then $V_t = U_{\rho(t)}$, and for $a \le s \le t \le b$ we have $\rho(t) \le \rho(s)$, so \begin{align*} V_t - V_s = U_{\rho(t)} - U_{\rho(s)} = -(U_{\rho(s)} - U_{\rho(t)}) = -\int_{\rho(t)}^{\rho(s)} U_\sigma \, d\overleftarrow{x}_\sigma. \end{align*} Substitute $\sigma = \rho(r) = a + b - r$ in the Young integral. As $r$ ranges over $[s, t]$, $\sigma = \rho(r)$ ranges over $[\rho(t), \rho(s)]$ in **reverse** order. The change-of-variable formula for Young integrals (which is valid for the Young integral with respect to a smooth and monotonic substitution) gives \begin{align*} \int_{\rho(t)}^{\rho(s)} U_\sigma \, d\overleftarrow{x}_\sigma = \int_{r=t}^{r=s} U_{\rho(r)} \, d\overleftarrow{x}_{\rho(r)}. \end{align*} Now, $\overleftarrow{x}_{\rho(r)} = x_{\rho(\rho(r))} = x_r$, since $\rho \circ \rho = \operatorname{id}$. Therefore $d\overleftarrow{x}_{\rho(r)} = dx_r$ as a Young differential. Thus \begin{align*} \int_{r=t}^{r=s} U_{\rho(r)} \, dx_r = \int_t^s V_r \, dx_r = -\int_s^t V_r \, dx_r, \end{align*} where the final sign comes from reversing the bounds of integration on the standard (forward) Young integral. Combining, \begin{align*} V_t - V_s = -\int_{\rho(t)}^{\rho(s)} U_\sigma \, d\overleftarrow{x}_\sigma = -\biggl(-\int_s^t V_r \, dx_r\biggr) = \int_s^t V_r \, dx_r. \end{align*} This is exactly the integral form of the CDE \begin{align*} dV_t = V_t \cdot dx_t, \end{align*} **driven by $x$**, but with the data specified at the terminal time $t = b$ rather than the initial time: \begin{align*} V_b = U_a = A. \end{align*} What about the sign? The substitution introduces a sign flip from $d\overleftarrow{x}_{\rho(r)} = dx_r$ (no sign) but the bound reversal contributes a sign. After bookkeeping, the two cancel exactly (since reversing time on $\overleftarrow{x}$ undoes the original time reversal that defined $\overleftarrow{x}$). The net result: $V$ satisfies the **forward** CDE with respect to $x$.[/guided]

[step:Apply uniqueness of the CDE driven by $x$ to identify $V_t = S(x)_{[a, t]}$ — but anchored at the terminal condition] We have shown $V$ solves $dV_t = V_t \cdot dx_t$ on $[a, b]$. We now claim that $V$ must equal $S(x)_{[a, \cdot]}$ — i.e. the standard signature path. The CDE $dY_t = Y_t \cdot dx_t$ with **initial** condition $Y_a = \mathbf{1}$ has, by [Signature as Solution of a CDE](/theorems/2497), the unique solution $Y_t = S(x)_{[a, t]}$. In particular, $Y_b = S(x)_{[a, b]} = A$. Both $V$ and $Y := S(x)_{[a, \cdot]}$ satisfy the same linear CDE \begin{align*} dZ_t = Z_t \cdot dx_t \quad \text{on } [a, b], \end{align*} and both achieve the same value $A$ at $t = b$ (we have $V_b = A$ from Step 2 and $Y_b = A$ by construction). Linear CDEs are uniquely determined by their value at any point in the interval (running backward from $b$, we get a uniquely-solvable linear backward CDE, by exactly the same Picard argument as in [Signature as Solution of a CDE](/theorems/2497) applied to the time-reversed driver). Therefore $V \equiv Y$ on $[a, b]$: \begin{align*} V_t = S(x)_{[a, t]} \qquad \text{for all } t \in [a, b]. \end{align*} In particular $V_a = S(x)_{[a, a]} = \mathbf{1}$.[/step]

[guided]The uniqueness statement of [Signature as Solution of a CDE](/theorems/2497) gives uniqueness for the **initial value problem**: given $Y_a$, $Y_t = Y_a \cdot S(x)_{[a, t]}$. We need uniqueness for the **terminal value problem**: given $V_b$, what is $V_t$? Two ways to see this gives uniqueness. First, **direct argument**: the equation \begin{align*} dZ_t = Z_t \cdot dx_t \quad \text{on } [a, b] \end{align*} is linear in $Z$. Linear CDEs admit a propagator: if $Z, \tilde Z$ are two solutions, then their difference $Z - \tilde Z$ also solves the same linear CDE (by linearity of the equation), and starts at $Z_a - \tilde Z_a$. By the existence-uniqueness theorem applied to the difference, $Z_t - \tilde Z_t = (Z_a - \tilde Z_a) \cdot S(x)_{[a, t]}$. Therefore if $Z_b = \tilde Z_b$, then $(Z_a - \tilde Z_a) \cdot S(x)_{[a, b]} = 0$. But $S(x)_{[a, b]}$ is invertible in $T((V))$ — its level-$0$ component is $1$, so it is in $\mathbf{1} + T_+((V))$, the unipotent group, which is closed under inversion. Therefore $Z_a = \tilde Z_a$, and consequently $Z = \tilde Z$ everywhere. (This is the standard argument that linear CDEs have unique solutions specified by any single point.) Second, **time-reversal argument**: the CDE $dZ_t = Z_t \cdot dx_t$ on $[a, b]$ with terminal value $Z_b$ specified can be re-cast as a forward CDE on the time-reversed interval driven by $\overleftarrow{x}$ with initial value $Z_b$. By [Signature as Solution of a CDE](/theorems/2497), this has a unique solution. Either way, $V$ is uniquely determined by $V_b = A$. We compare with $Y_t := S(x)_{[a, t]}$, which satisfies the **same** CDE (by [Signature as Solution of a CDE](/theorems/2497) with $A = \mathbf{1}$) and has $Y_b = S(x)_{[a, b]} = A$. So $V$ and $Y$ are two solutions of the same linear CDE with the same value at $b$; uniqueness forces \begin{align*} V_t = Y_t = S(x)_{[a, t]} \qquad \text{for all } t \in [a, b]. \end{align*} In particular, \begin{align*} V_a = S(x)_{[a, a]} = \mathbf{1}. \end{align*} This is the key consequence: the value of $V$ at the **start** of $[a, b]$ is $\mathbf{1}$.[/guided]

[step:Translate $V_a = \mathbf{1}$ back to the original variables and conclude]By construction in Step 2, $V_a = U_b$. By Step 1, $U_b = A \cdot S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]}$. By Step 3, $V_a = \mathbf{1}$. Combining, \begin{align*} S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]} = \mathbf{1}. \end{align*} For the other identity, observe that $\overleftarrow{\overleftarrow{x}}_t = \overleftarrow{x}_{a + b - t} = x_{a + b - (a + b - t)} = x_t$, so $\overleftarrow{\overleftarrow{x}} = x$. Apply the identity already proved with $x$ replaced by $\overleftarrow{x}$: \begin{align*} S(\overleftarrow{x})_{[a, b]} \cdot S(\overleftarrow{\overleftarrow{x}})_{[a, b]} = \mathbf{1}, \end{align*} i.e. \begin{align*} S(\overleftarrow{x})_{[a, b]} \cdot S(x)_{[a, b]} = \mathbf{1}. \end{align*} These two identities together state that $S(\overleftarrow{x})_{[a, b]}$ is both a left and a right inverse of $S(x)_{[a, b]}$ in $T((V))$. In a (possibly non-commutative) associative unital algebra, having both a left and a right inverse implies they coincide and the element is invertible with the unique inverse being this common value. Therefore \begin{align*} S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]}^{-1}, \end{align*} completing the proof.[/step]

[guided]We assemble the pieces. From Step 2: $V_a = U_b$ (definition of $V$). From Step 1: $U_b = A \cdot S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]}$ (the explicit formula for the solution of the CDE driven by $\overleftarrow{x}$). From Step 3: $V_a = \mathbf{1}$ (uniqueness of the linear CDE driven by $x$, with terminal value $A$). Putting these together: \begin{align*} S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]} = U_b = V_a = \mathbf{1}. \end{align*} This is the **first** half of the claimed identity. For the **second** half: $S(\overleftarrow{x})_{[a, b]} \cdot S(x)_{[a, b]} = \mathbf{1}$. There are two clean ways to derive this. **Option 1 — Use $\overleftarrow{\overleftarrow{x}} = x$.** Replace $x$ by $\overleftarrow{x}$ in the identity just proved. The hypothesis is symmetric: $\overleftarrow{x} \in C_p([a, b], V)$ since $C_p$ is symmetric under time reversal. Applying the result, \begin{align*} S(\overleftarrow{x})_{[a, b]} \cdot S(\overleftarrow{\overleftarrow{x}})_{[a, b]} = \mathbf{1}. \end{align*} And $\overleftarrow{\overleftarrow{x}}_t = x_t$ (time-reversing twice gives back the original path), so $S(\overleftarrow{\overleftarrow{x}}) = S(x)$. Hence \begin{align*} S(\overleftarrow{x})_{[a, b]} \cdot S(x)_{[a, b]} = \mathbf{1}. \end{align*} **Option 2 — Algebraic.** We have shown $S(x)_{[a, b]} \cdot S(\overleftarrow{x})_{[a, b]} = \mathbf{1}$, which says $S(\overleftarrow{x})_{[a, b]}$ is a **right inverse** of $S(x)_{[a, b]}$ in $T((V))$. We want to upgrade this to a two-sided inverse. The element $S(x)_{[a, b]} \in \mathbf{1} + T_+((V))$ (its level-$0$ component is $1$), and any element of this form is invertible in $T((V))$ — in fact the inverse is given by the geometric series $\sum_{m \ge 0} (-1)^m (S(x)_{[a, b]} - \mathbf{1})^{\otimes m}$, which converges level-wise. The two-sided inverse coincides with the right inverse, so $S(\overleftarrow{x})_{[a, b]}$ is also a left inverse. Either way, $S(\overleftarrow{x})_{[a, b]} \cdot S(x)_{[a, b]} = \mathbf{1}$. Together, the two identities show that $S(x)_{[a, b]}$ is invertible in $T((V))$ with inverse $S(\overleftarrow{x})_{[a, b]}$, i.e. \begin{align*} S(\overleftarrow{x})_{[a, b]} = S(x)_{[a, b]}^{-1}. \end{align*} The result is intuitively obvious — running a path forward and then back should leave nothing — but the proof reveals that this is not an algebraic accident: it is a consequence of the **uniqueness** of CDE solutions, which is a genuinely analytic input.[/guided]