[proofplan] We prove the identity at every level $k$ of the signature. The argument is established first in the bounded-variation case $p = 1$, where iterated integrals are classical Stieltjes integrals: a level-$k$ iterated integral over $[a, c]$ decomposes into a sum, indexed by $i = 0, 1, \dots, k$, of integrals over the simplex $\{a < u_1 < \cdots < u_i < b < u_{i+1} < \cdots < u_k < c\}$. Each summand factorises by Fubini's theorem into a tensor product of a level-$i$ signature on $[a, b]$ and a level-$(k-i)$ signature on $[b, c]$, which is exactly the level-$k$ component of the tensor-algebra product $S(x) \cdot S(y)$. The case $1 < p < 2$ then follows by approximation: bounded-variation paths are dense in $q$-variation for any $q > 1$, and the signature map and tensor-algebra product are continuous in the $q$-variation topology when $1/p + 1/q > 1$. [/proofplan] [step:Set up the concatenated path $z = x \ast y$ and its level-$k$ iterated integral] Recall the [concatenation operation](/page/Path%20Concatenation): for $x \in C_p([a, b], V)$ and $y \in C_p([b, c], V)$ with $x_b = y_b$ (assumed implicitly via the standard convention that concatenation pre-translates $y$ so the endpoint matches the starting point of the second segment, or via the assumption that $x_b$ matches $y_b$ after any required translation), the concatenation \begin{align*} z := x \ast y: [a, c] &\to V \\ t &\mapsto \begin{cases} x_t & \text{if } t \in [a, b], \\ y_t & \text{if } t \in [b, c]. \end{cases} \end{align*} The path $z$ is continuous on $[a, c]$ by the matching condition at $t = b$. The product on the tensor algebra $T((V))$ acts component-wise as \begin{align*} \bigl(S(x) \cdot S(y)\bigr)^{(k)} = \sum_{i=0}^k S(x)^{(i)} \otimes S(y)^{(k-i)} \in V^{\otimes k}, \end{align*} with the conventions $S(x)^{(0)} = S(y)^{(0)} = 1$ and $V^{\otimes 0} = \mathbb{R}$ (so the level-zero terms in the sum contribute elementary tensor factors). To prove $S(z)_{[a, c]} = S(x)_{[a, b]} \cdot S(y)_{[b, c]}$ in $T((V))$, it suffices to prove \begin{align*} S(z)^{(k)}_{[a,c]} = \sum_{i=0}^k S(x)^{(i)}_{[a,b]} \otimes S(y)^{(k-i)}_{[b,c]} \qquad \forall k \ge 0. \end{align*} The case $k = 0$ is the identity $1 = 1$. For $k \ge 1$ we proceed by direct computation. [/step] [step:Establish the case $p = 1$ by splitting the simplex of integration] Assume $p = 1$, so $x, y, z$ are all of bounded variation. Coordinate-wise, fix a basis $(e_1, \dots, e_d)$ of $V$ and a multi-index $(j_1, \dots, j_k) \in \{1, \dots, d\}^k$. By definition of the iterated integral, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a, c]} = \int_{a < u_1 < u_2 < \cdots < u_k < c} dz^{j_1}_{u_1} \cdots dz^{j_k}_{u_k}, \end{align*} interpreted iteratively in the order $u_1$ innermost, $u_k$ outermost. The Lebesgue–Stieltjes measure $\mu^j_z$ on $[a, c]$ associated to $z^j$ satisfies \begin{align*} \mu^j_z\big|_{[a, b]} = \mu^j_x, \qquad \mu^j_z\big|_{[b, c]} = \mu^j_y, \qquad \mu^j_z(\{b\}) = 0 \end{align*} (the third statement holds because $z^j$ is continuous at $b$, so the Stieltjes measure has no atom there). Therefore the integration variables $u_1, \dots, u_k$ effectively partition into those in $[a, b)$ and those in $(b, c]$ (the single point $b$ contributing nothing to any integral). The integration domain $\{a < u_1 < u_2 < \cdots < u_k < c\}$ decomposes, up to a Lebesgue/Stieltjes-null subset (the boundary $\{u_i = b\}$ for some $i$), as the disjoint union \begin{align*} \bigsqcup_{i=0}^k \bigl\{a < u_1 < \cdots < u_i < b < u_{i+1} < \cdots < u_k < c\bigr\}, \end{align*} where the $i = 0$ term is $\{b < u_1 < \cdots < u_k < c\}$ and the $i = k$ term is $\{a < u_1 < \cdots < u_k < b\}$. (The decomposition follows from the chain $a < u_1 < \cdots < u_k < c$ and the cut at $b$: there is some unique index $i \in \{0, 1, \dots, k\}$ with $u_i < b < u_{i+1}$, where by convention $u_0 := a$ and $u_{k+1} := c$.) Splitting the iterated integral along this decomposition and using the additivity of Stieltjes integrals (and the absence of mass at $\{b\}$), \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dz^{j_1}_{u_1} \cdots dz^{j_i}_{u_i} \, dz^{j_{i+1}}_{u_{i+1}} \cdots dz^{j_k}_{u_k}. \end{align*} On the first set of variables (those $\le b$) the measure $\mu^{j_\ell}_z$ agrees with $\mu^{j_\ell}_x$; on the second set (those $\ge b$) it agrees with $\mu^{j_\ell}_y$. Substituting, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \, dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}. \end{align*} [guided] The strategy. The signature of the concatenation $z = x \ast y$ is, by definition, an iterated integral over the ordered simplex $\{a < u_1 < \cdots < u_k < c\}$ inside the larger interval $[a, c]$. The path $z$ has different defining rules on the two halves of $[a, c]$: $z|_{[a,b]} = x$ and $z|_{[b,c]} = y$. We exploit this by classifying the integration variables according to which half they fall in, and splitting the simplex accordingly. Setting up. Coordinate-wise, fix a basis $(e_1, \dots, e_d)$ of $V$ and a multi-index $(j_1, \dots, j_k) \in \{1, \dots, d\}^k$. By definition of the iterated integral, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a, c]} = \int_{a < u_1 < u_2 < \cdots < u_k < c} dz^{j_1}_{u_1} \cdots dz^{j_k}_{u_k}, \end{align*} the Stieltjes integration interpreted iteratively from $u_1$ (innermost) to $u_k$ (outermost). Identifying the relevant Stieltjes measures. The Lebesgue–Stieltjes measure $\mu^j_z$ on $[a, c]$ associated to the coordinate $z^j$ has three properties on the cut interval: \begin{align*} \mu^j_z\big|_{[a, b]} = \mu^j_x, \qquad \mu^j_z\big|_{[b, c]} = \mu^j_y, \qquad \mu^j_z(\{b\}) = 0. \end{align*} The first two follow from the definition $z^j_t = x^j_t$ for $t \in [a,b]$ and $z^j_t = y^j_t$ for $t \in [b,c]$, applied to the half-open generators. The third (no atom at $b$) holds because $z^j$ is continuous at $b$ by construction of the concatenation: $z^j_b = x^j_b = y^j_b$ by the matching condition. A continuous Stieltjes integrator yields a non-atomic Stieltjes measure, so single points have zero mass. Splitting the simplex. The integration domain $\{a < u_1 < u_2 < \cdots < u_k < c\}$ decomposes, up to a $\mu^j_z$-null set, as the disjoint union \begin{align*} \bigsqcup_{i=0}^k \bigl\{a < u_1 < \cdots < u_i < b < u_{i+1} < \cdots < u_k < c\bigr\}, \end{align*} where the $i = 0$ piece is $\{b < u_1 < \cdots < u_k < c\}$ and the $i = k$ piece is $\{a < u_1 < \cdots < u_k < b\}$. Why is this decomposition correct? Because for any point $(u_1, \dots, u_k)$ in the simplex with $u_\ell \ne b$ for all $\ell$ (a co-null condition), the chain $u_1 < \cdots < u_k$ together with the cut at $b$ forces the existence of a **unique** index $i \in \{0, 1, \dots, k\}$ with $u_i < b < u_{i+1}$ (with the conventions $u_0 := a < b$ and $u_{k+1} := c > b$). The points where some $u_\ell = b$ form a $\mu^j_z$-null set by the no-atom-at-$b$ property, so we can ignore them in the integral. Splitting the integral. By countable additivity of Stieltjes integrals over disjoint domains (and the absence of mass at $\{u_\ell = b\}$), \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dz^{j_1}_{u_1} \cdots dz^{j_i}_{u_i} \, dz^{j_{i+1}}_{u_{i+1}} \cdots dz^{j_k}_{u_k}. \end{align*} Substituting back $z = x$ and $z = y$. On the first set of variables (those in $[a, b]$), the measure $\mu^{j_\ell}_z$ agrees with $\mu^{j_\ell}_x$. On the second set (those in $[b, c]$), it agrees with $\mu^{j_\ell}_y$. Substituting, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \, dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}. \end{align*} This is the Chen identity in nascent form: the level-$k$ simplex over $[a, c]$ has been broken into $k+1$ disjoint pieces, each corresponding to a specific split of the multi-index between $x$ on $[a,b]$ and $y$ on $[b,c]$. The next step (Step 3) factorises each summand via Fubini into a tensor product of a level-$i$ signature of $x$ and a level-$(k-i)$ signature of $y$, completing the proof. Why is the no-atom-at-$b$ property essential? If $z^j$ were discontinuous at $b$ (i.e., $x_b \ne y_b$ before any matching translation), the Stieltjes measure $\mu^j_z$ would have a Dirac mass at $b$, and the decomposition would acquire an extra "boundary" piece corresponding to $u_\ell = b$. This would not produce the clean tensor-product factorisation we need. The matching condition $x_b = y_b$ is precisely what ensures Chen's identity holds in clean form; without it, one would get a corrected version with boundary jumps. [/guided] [/step] [step:Apply Fubini's theorem to factorise each summand into a tensor product] Fix $i \in \{0, 1, \dots, k\}$ and consider the $i$-th summand: \begin{align*} T_i := \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \, dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}. \end{align*} The integration domain is the Cartesian product \begin{align*} \Delta_i^{[a,b]} \times \Delta_{k-i}^{[b,c]} \subseteq [a, b]^i \times [b, c]^{k-i}, \end{align*} where $\Delta_i^{[a,b]} := \{(u_1, \dots, u_i) \in [a, b]^i : a < u_1 < \cdots < u_i < b\}$ and analogously for $\Delta_{k-i}^{[b,c]}$. The integrand is the product of an $x$-dependent factor (involving only the variables $u_1, \dots, u_i$) and a $y$-dependent factor (involving only $u_{i+1}, \dots, u_k$): \begin{align*} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \otimes dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}. \end{align*} Apply [Fubini's theorem](/theorems/???) to factorise the integral over the product domain into a product of two iterated integrals. Fubini's hypothesis (integrability with respect to the product measure) is satisfied because both factor integrals are finite — they are themselves bounded variation iterates, controlled by [Factorial Decay](/theorems/2493) (when $p = 1$) which guarantees $\|S(x)^{(i)}\| \le \|x\|_{1;[a,b]}^i / i! < \infty$ and likewise for $y$. Factorising, \begin{align*} T_i = \left( \int_{a < u_1 < \cdots < u_i < b} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \right) \otimes \left( \int_{b < u_{i+1} < \cdots < u_k < c} dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k} \right). \end{align*} The two factors are exactly $S(x)^{(j_1, \dots, j_i)}_{[a, b]}$ and $S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}$. Hence \begin{align*} T_i = S(x)^{(j_1, \dots, j_i)}_{[a, b]} \otimes S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}. \end{align*} Substituting back into the formula from Step 2, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k S(x)^{(j_1, \dots, j_i)}_{[a, b]} \otimes S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}. \end{align*} Re-indexing on the right (with $i$ ranging over level-splits and the multi-indices summed implicitly along their split), and assembling across all multi-indices on the basis $(e_{j_1} \otimes \cdots \otimes e_{j_k})$ of $V^{\otimes k}$, \begin{align*} S(z)^{(k)}_{[a, c]} = \sum_{i=0}^k S(x)^{(i)}_{[a,b]} \otimes S(y)^{(k-i)}_{[b,c]} = \bigl(S(x)_{[a,b]} \cdot S(y)_{[b,c]}\bigr)^{(k)}. \end{align*} Since this holds for every $k \ge 0$, we have $S(z)_{[a, c]} = S(x)_{[a, b]} \cdot S(y)_{[b, c]}$ in $T((V))$, proving the case $p = 1$. [guided] The strategy. Each summand $T_i$ from Step 2 is a Stieltjes integral over a Cartesian product of two simplices, with an integrand that factors as a tensor product of an $x$-dependent factor and a $y$-dependent factor. We use [Fubini's theorem](/theorems/???) to split the integral into the product of two simpler integrals, each of which is recognised as a signature component. Identifying the product structure. Fix $i \in \{0, 1, \dots, k\}$ and consider \begin{align*} T_i = \int_{\substack{a < u_1 < \cdots < u_i < b \\ b < u_{i+1} < \cdots < u_k < c}} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \, dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}. \end{align*} The integration domain is a Cartesian product: \begin{align*} \Delta_i^{[a,b]} \times \Delta_{k-i}^{[b,c]} \subseteq [a, b]^i \times [b, c]^{k-i}, \end{align*} where $\Delta_i^{[a,b]} := \{(u_1, \dots, u_i) \in [a,b]^i : a < u_1 < \cdots < u_i < b\}$ is the $i$-th ordered simplex inside $[a,b]^i$, and analogously $\Delta_{k-i}^{[b,c]}$. The two simplex factors involve **disjoint** sets of variables ($u_1, \dots, u_i$ versus $u_{i+1}, \dots, u_k$) and there are **no constraints** linking the two groups — the cut at $b$ is already encoded in the separate domains, not in any joint inequality. So the integration is genuinely over a product set. The integrand factorises in tandem: the variables $u_1, \dots, u_i$ appear only in the $x$-differentials $dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i}$, and $u_{i+1}, \dots, u_k$ appear only in the $y$-differentials $dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}$. Symbolically, \begin{align*} \text{(integrand)} = \bigl(dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i}\bigr) \otimes \bigl(dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}\bigr), \end{align*} where $\otimes$ marks the tensor factorisation in $V^{\otimes i} \otimes V^{\otimes (k-i)} = V^{\otimes k}$ (Recall the basis decomposition: each $dx^{j_\ell}_{u_\ell}$ is the $e_{j_\ell}$-component of $dx_{u_\ell}$.) Verifying Fubini's hypothesis. To apply [Fubini's theorem](/theorems/???) (or its product-of-finite-measures variant for Stieltjes integrals) we need integrability of the integrand against the product measure $\mu^{j_1}_x \otimes \cdots \otimes \mu^{j_i}_x \otimes \mu^{j_{i+1}}_y \otimes \cdots \otimes \mu^{j_k}_y$ on the product domain. Each factor integral is finite by [Factorial Decay](/theorems/2493): for $p = 1$, \begin{align*} \bigl|S(x)^{(j_1, \dots, j_i)}_{[a, b]}\bigr| \le \frac{\|x\|_{1;[a,b]}^i}{i!} < \infty, \qquad \bigl|S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}\bigr| \le \frac{\|y\|_{1;[b,c]}^{k-i}}{(k-i)!} < \infty. \end{align*} The product of two finite-mass signed measures has finite total variation on the product domain, and the integrand $\mathbb{1}_{\Delta_i \times \Delta_{k-i}}$ is bounded, so Fubini applies in the variation-controlled Stieltjes sense. Factorising. Applying Fubini, \begin{align*} T_i &= \int_{\Delta_i^{[a,b]} \times \Delta_{k-i}^{[b,c]}} \bigl(dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i}\bigr) \otimes \bigl(dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k}\bigr) \\ &= \left( \int_{a < u_1 < \cdots < u_i < b} dx^{j_1}_{u_1} \cdots dx^{j_i}_{u_i} \right) \otimes \left( \int_{b < u_{i+1} < \cdots < u_k < c} dy^{j_{i+1}}_{u_{i+1}} \cdots dy^{j_k}_{u_k} \right). \end{align*} The two factors are exactly $S(x)^{(j_1, \dots, j_i)}_{[a,b]}$ and $S(y)^{(j_{i+1}, \dots, j_k)}_{[b,c]}$ by the recursive definition of iterated integrals (using the conventions $S(x)^{()}_{[a,b]} = 1$ at level $0$, which handles the boundary cases $i = 0$ and $i = k$). Hence \begin{align*} T_i = S(x)^{(j_1, \dots, j_i)}_{[a, b]} \otimes S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}. \end{align*} Assembling. Substituting back into the formula from Step 2, \begin{align*} S(z)^{(j_1, \dots, j_k)}_{[a,c]} = \sum_{i=0}^k S(x)^{(j_1, \dots, j_i)}_{[a, b]} \otimes S(y)^{(j_{i+1}, \dots, j_k)}_{[b, c]}. \end{align*} Now sum over all multi-indices $(j_1, \dots, j_k) \in \{1, \dots, d\}^k$ with the basis tensor $e_{j_1} \otimes \cdots \otimes e_{j_k}$ of $V^{\otimes k}$. The right-hand side, when re-aggregated, equals the level-$k$ component of the tensor-algebra product: \begin{align*} S(z)^{(k)}_{[a, c]} = \sum_{i=0}^k S(x)^{(i)}_{[a,b]} \otimes S(y)^{(k-i)}_{[b,c]} = \bigl(S(x)_{[a,b]} \cdot S(y)_{[b,c]}\bigr)^{(k)}. \end{align*} This holds for every $k \ge 0$, so $S(z)_{[a, c]} = S(x)_{[a, b]} \cdot S(y)_{[b, c]}$ in $T((V))$, completing the case $p = 1$. The big picture. Chen's identity is an instance of a homomorphism property: the signature map $S: (\text{paths}, \ast) \to (T((V)), \cdot)$ takes concatenation of paths to the tensor-algebra product. The combinatorics underlying this homomorphism is exactly the **deconcatenation coproduct** in the tensor algebra dual to the shuffle algebra: the $k+1$ ways to split a multi-index $(j_1, \dots, j_k)$ into a prefix $(j_1, \dots, j_i)$ and a suffix $(j_{i+1}, \dots, j_k)$ correspond to the $k+1$ ways the integration variables can fall on either side of $b$. [/guided] [/step] [step:Extend to $1 < p < 2$ by density and continuity of Young integration] Now suppose $1 < p < 2$. Choose $q > 1$ with $1/p + 1/q > 1$ and $q < 2$ (possible because $1/p > 1/2$, so $1/q$ can be chosen in the non-empty interval $(1/2, 1 - 1/p) \cup (\max(0, 1 - 1/p), 1)$; concretely any $q$ with $\max(1, p/(p-1) - \varepsilon) > q > 1$ for small $\varepsilon > 0$ works). By the [density of bounded-variation paths in $q$-variation](/theorems/???), there exist sequences $(x^{(n)})_{n \in \mathbb{N}} \subset C_1([a, b], V)$ and $(y^{(n)})_{n \in \mathbb{N}} \subset C_1([b, c], V)$ with \begin{align*} \|x - x^{(n)}\|_{q;[a,b]} \to 0, \qquad \|y - y^{(n)}\|_{q;[b,c]} \to 0. \end{align*} Choose the approximations to additionally satisfy the matching condition $x^{(n)}_b = y^{(n)}_b$ (this can be arranged by translating each $y^{(n)}$ by the small vector $x^{(n)}_b - y^{(n)}_b \to 0$, which preserves the $q$-variation since translation is invariant). Then $z^{(n)} := x^{(n)} \ast y^{(n)} \in C_1([a, c], V)$ is well-defined, and \begin{align*} \|z - z^{(n)}\|_{q;[a,c]} \to 0 \end{align*} (the $q$-variation of a concatenated path on $[a, c]$ is controlled by the sum of $q$-variations on $[a, b]$ and $[b, c]$ plus the matching jump, which goes to zero). Two continuity facts apply: (a) The signature map $S: C_q \to T((V))$ is **jointly continuous in $q$-variation** when $1/p + 1/q > 1$. This follows from the [continuity of the Young integral](/theorems/???) applied iteratively: at each level $k$, the map $(f, g) \mapsto \int f \, dg$ is jointly continuous in $q$-variation, and signatures are built by composing such bilinear operations. (b) The tensor-algebra product $\cdot: T((V)) \times T((V)) \to T((V))$ is continuous level-by-level in any reasonable topology on $T((V))$ that controls finitely many levels: at level $k$, the product $(A \cdot B)^{(k)} = \sum_{i=0}^k A^{(i)} \otimes B^{(k-i)}$ is a finite sum of bilinear maps on $V^{\otimes i} \times V^{\otimes (k-i)}$, hence continuous in any norm on each homogeneous component. By Step 3 applied to the bounded-variation paths $x^{(n)}, y^{(n)}, z^{(n)}$, \begin{align*} S(z^{(n)})_{[a, c]} = S(x^{(n)})_{[a, b]} \cdot S(y^{(n)})_{[b, c]} \qquad \forall n. \end{align*} Pass to the limit. By (a), $S(z^{(n)}) \to S(z)$, $S(x^{(n)}) \to S(x)$, $S(y^{(n)}) \to S(y)$ in the $q$-variation topology on $T((V))$ at every level. By (b), the right-hand side converges to $S(x) \cdot S(y)$ at every level. Equality of the limits gives \begin{align*} S(z)_{[a, c]} = S(x)_{[a, b]} \cdot S(y)_{[b, c]}, \end{align*} completing the proof. [guided] The strategy. Prove Chen's identity on a dense subclass where it is easy ($p = 1$, Step 3), then transport it to the full class by continuity. The dense subclass is $C_1$ (bounded-variation paths), where iterated integrals are honest Stieltjes integrals and the simplex-splitting argument from Steps 2–3 applies directly. To make the transport work we need (i) approximating sequences in a suitable topology in which (ii) both sides of Chen's identity are continuous functions of the inputs. Choosing the approximating topology. We choose the $q$-variation topology with $q > 1$ satisfying $1/p + 1/q > 1$ and $q < 2$. Such $q$ exists because the constraint $1/p + 1/q > 1$ becomes $1/q > 1 - 1/p$, i.e., $q < p/(p-1)$. Since $p < 2$, we have $p/(p-1) > 2$, so we can choose any $q \in (1, 2)$ with $q < p/(p-1)$. Constructing the approximating sequences. By the [density of bounded-variation paths in $q$-variation](/theorems/???), \begin{align*} \exists (x^{(n)})_{n \in \mathbb{N}} \subset C_1([a, b], V), \quad &\|x - x^{(n)}\|_{q;[a,b]} \to 0, \\ \exists (y^{(n)})_{n \in \mathbb{N}} \subset C_1([b, c], V), \quad &\|y - y^{(n)}\|_{q;[b,c]} \to 0. \end{align*} Matching condition. We need the concatenations $z^{(n)} := x^{(n)} \ast y^{(n)}$ to be well-defined, which requires $x^{(n)}_b = y^{(n)}_b$. After approximating, the matching may fail by a small amount $\delta_n := x^{(n)}_b - y^{(n)}_b \in V$. Since $x^{(n)} \to x$ and $y^{(n)} \to y$ in $q$-variation (which controls supremum), and $x_b = y_b$ by hypothesis, we have $\delta_n \to 0$ in $V$. We translate each $y^{(n)}$ by $\delta_n$, redefining $\tilde y^{(n)} := y^{(n)} + \delta_n$ (a constant translation). This translation: (a) preserves $q$-variation (translation by a constant is invariant under finite-difference operators), so $\|\tilde y^{(n)} - y\|_{q;[b,c]} \le \|y^{(n)} - y\|_{q;[b,c]} + |\delta_n| \to 0$; (b) restores the matching $x^{(n)}_b = \tilde y^{(n)}_b$. Replacing $y^{(n)}$ by $\tilde y^{(n)}$ from now on, the concatenations $z^{(n)} := x^{(n)} \ast \tilde y^{(n)} \in C_1([a, c], V)$ are well-defined. Convergence of the concatenations. The $q$-variation of a concatenated path on $[a, c]$ is bounded by the sum of $q$-variations on the two halves plus a matching-jump term that vanishes when the matching holds: \begin{align*} \|z - z^{(n)}\|_{q;[a,c]} \le \|x - x^{(n)}\|_{q;[a,b]} + \|y - \tilde y^{(n)}\|_{q;[b,c]} + (\text{matching-jump term, $\to 0$}) \to 0. \end{align*} Two continuity facts. We need both sides of Chen's identity to be continuous in the $q$-variation topology. (a) **Signature continuity**: the signature map $S: C_q \to T((V))$ is jointly continuous in $q$-variation when $1/p + 1/q > 1$. This is the [continuity of the Young integral](/theorems/???) applied iteratively level by level: at each level $k$, the bilinear map $(f, g) \mapsto \int f \, dg$ is jointly continuous in the relevant $q$-variation seminorms, and the iterated integral $S(\cdot)^{(k)}$ is built by composing $k$ such bilinear operations. Continuity of compositions of continuous maps gives continuity of $S$ at every level. (b) **Tensor-product continuity**: the tensor-algebra product $\cdot: T((V)) \times T((V)) \to T((V))$ is continuous level-by-level. At level $k$, the formula \begin{align*} (A \cdot B)^{(k)} = \sum_{i=0}^k A^{(i)} \otimes B^{(k-i)} \end{align*} is a finite sum of bilinear maps $V^{\otimes i} \times V^{\otimes (k-i)} \to V^{\otimes k}$. Each of these is continuous in any norm on the homogeneous components (when $V$ is finite-dimensional, this is automatic; for general $V$, it follows from the universal property of the tensor product applied to bounded bilinear maps). Passing to the limit. By Step 3 applied to the bounded-variation triple $(x^{(n)}, \tilde y^{(n)}, z^{(n)})$, \begin{align*} S(z^{(n)})_{[a, c]} = S(x^{(n)})_{[a, b]} \cdot S(\tilde y^{(n)})_{[b, c]} \qquad \forall n. \end{align*} Using (a) on each side: $S(z^{(n)})_{[a,c]} \to S(z)_{[a,c]}$, $S(x^{(n)})_{[a,b]} \to S(x)_{[a,b]}$, $S(\tilde y^{(n)})_{[b,c]} \to S(y)_{[b,c]}$ at every level. Using (b): the right-hand side $S(x^{(n)})_{[a,b]} \cdot S(\tilde y^{(n)})_{[b,c]}$ converges to $S(x)_{[a,b]} \cdot S(y)_{[b,c]}$ at every level. Equality of limits gives Chen's identity: \begin{align*} S(z)_{[a, c]} = \lim_n S(z^{(n)})_{[a, c]} = \lim_n S(x^{(n)})_{[a, b]} \cdot S(\tilde y^{(n)})_{[b, c]} = S(x)_{[a, b]} \cdot S(y)_{[b, c]}. \end{align*} This completes the proof. Why $p < 2$? The Young-integration regime requires $1/p + 1/q > 1$ with $q > 1$, equivalently $p < q/(q-1)$. The right-hand side decreases from $\infty$ (at $q = 1^+$) to $1$ (at $q \to \infty$), passing through $2$ at $q = 2$. So for $p < 2$ we can find $q > 1$ with $q/(q-1) > p$ and $q < 2$, but for $p \ge 2$ no such $q > 1$ exists in the Young regime. The threshold $p = 2$ is the boundary of Young integration; for $p \ge 2$, Chen's identity still holds but requires the rough-path lift to define iterated integrals in the first place. [/guided] [/step]