[proofplan] A topological group is a group with continuous multiplication and continuous inversion. The group structure on $\mathcal{C}_p$ is inherited from the group of group-like elements $T_1((V))$ in the tensor algebra via the signature map $S$, which is a homeomorphism onto $\mathcal{S}_p \subset T_1((V))$ for the projective topology $\chi_{\mathrm{pr}}$. The strategy is therefore: factor inversion as $[\,\cdot\,]^{-1} = S^{-1} \circ \psi \circ S$ where $\psi$ is the algebraic inverse on $T_1((V))$, and factor multiplication analogously; then verify continuity at the $T_1((V))$ level using the universal property of the product topology — it suffices to check continuity level by level, where each level reduces to a finite multilinear computation. [/proofplan]

[step:Verify the group axioms]The group operations $[x] \cdot [y] = [x \ast y]$ and $[x]^{-1} = [\overleftarrow{x}]$ on $\mathcal{C}_p$ correspond, via the signature map $S$, to multiplication and inversion in the group $(T_1((V)), \otimes)$ of group-like elements of the tensor algebra: by [Chen's identity for signatures](/theorems/???), \begin{align*} S(x \ast y) = S(x) \otimes S(y), \qquad S(\overleftarrow{x}) = S(x)^{-1}, \end{align*} with the second identity meaning $S(\overleftarrow{x}) \otimes S(x) = \mathbf{1} = S(x) \otimes S(\overleftarrow{x})$. Since $S$ is injective on $\mathcal{C}_p$ and $T_1((V))$ is a group under $\otimes$, the operations descend to make $\mathcal{C}_p$ a group with identity $[\mathbf{1}]$ corresponding to $\mathbf{1} \in T_1((V))$. We focus the remainder of the proof on continuity.[/step]

[guided]A topological group needs three things: a group structure, a topology, and continuity of the group operations. The topology $\chi_{\mathrm{pr}}$ is given. The group operations are also given by definition — concatenation and reversal of paths, descending to $\mathcal{C}_p$ — but to use them confidently we should convince ourselves they really do define a group, with all axioms (associativity, identity, inverses) satisfied. The cleanest way to verify the axioms is to *transport them from $T_1((V))$*, the group of group-like elements of the tensor algebra, where they are already known. The bridge is the signature map $S : \mathcal{C}_p \to T_1((V))$, which is a bijection onto $\mathcal{S}_p$ and intertwines the group operations. **Chen's identity.** The two key intertwining relations are \begin{align*} S(x \ast y) &= S(x) \otimes S(y), \\ S(\overleftarrow{x}) &= S(x)^{-1}, \end{align*} where $\ast$ denotes concatenation of paths and $\overleftarrow{x}$ the time-reversal of $x$. The first is [Chen's identity for signatures](/theorems/???) and follows directly from splitting an iterated integral over a concatenated path into pieces over each segment; the second says reversal corresponds to algebraic inversion in $T_1((V))$. Both identities are independent of the chosen representative of $[x]$ (they hold up to tree-equivalence), so they descend cleanly to $\mathcal{C}_p$. **Group axioms via transport.** $(T_1((V)), \otimes)$ is a group: associativity of $\otimes$ holds in any algebra, the multiplicative identity $\mathbf{1} \in T_1((V))$ is the unit, and Step 3 below will explicitly construct two-sided inverses. The bijection $S$ pulls these axioms back to $\mathcal{C}_p$: - *Associativity.* For $[x], [y], [z] \in \mathcal{C}_p$, $S(([x] \cdot [y]) \cdot [z]) = (S(x) \otimes S(y)) \otimes S(z) = S(x) \otimes (S(y) \otimes S(z)) = S([x] \cdot ([y] \cdot [z]))$. By injectivity of $S$ on $\mathcal{C}_p$, $([x] \cdot [y]) \cdot [z] = [x] \cdot ([y] \cdot [z])$. - *Identity.* The constant path $\mathbf{0}$ (or any tree-like path) has signature $\mathbf{1}$; for any $x$, $S([x] \cdot [\mathbf{0}]) = S(x) \otimes \mathbf{1} = S(x)$, so $[x] \cdot [\mathbf{0}] = [x]$. Similarly on the left. - *Inverses.* $S([x] \cdot [\overleftarrow{x}]) = S(x) \otimes S(x)^{-1} = \mathbf{1} = S([\mathbf{0}])$, so $[x] \cdot [\overleftarrow{x}] = [\mathbf{0}]$. Similarly on the other side. **Why we can focus on continuity.** The group axioms are now verified. The remaining content of "topological group" is *continuity* of multiplication and inversion. The factoring strategy that comes next — through the homeomorphism $S$ — is exactly the strategy this section motivates: argue group-theoretic structure on $T_1((V))$, then transport via $S$, including continuity once we establish it.[/guided]

[step:Factor inversion through the signature map]The signature map \begin{align*} S : (\mathcal{C}_p, \chi_{\mathrm{pr}}) &\to T_1((V)) \subset T((V)), \\ [x] &\mapsto S(x) \end{align*} is a homeomorphism onto its image $\mathcal{S}_p$. The topology $\chi_{\mathrm{pr}}$ is by construction the initial topology on $\mathcal{C}_p$ generated by the family $\{\pi_n \circ S : n \ge 0\}$, so $S$ is continuous (each generator factors through it). The inverse $S^{-1}: \mathcal{S}_p \to \mathcal{C}_p$ is continuous because, equipping $\mathcal{S}_p$ with the subspace topology from the product topology on $T((V))$, the open sets of $\chi_{\mathrm{pr}}$ on $\mathcal{C}_p$ are exactly the preimages under $S$ of open sets of the form $S^{-1}(\bigcap_{n \in F}(\pi_n)^{-1}(U_n))$ for finite $F$ and open $U_n \subset V^{\otimes n}$ — that is, $S$ is a homeomorphism onto its image by the very definition of $\chi_{\mathrm{pr}}$ as the initial topology. On $T((V))$ we use the product topology induced by the level projections $\pi_n: T((V)) \to V^{\otimes n}$. For $a \in T_1((V))$ — i.e. $\pi_0(a) = 1$ — write $a = \mathbf{1} + b$ where $b := a - \mathbf{1}$ has zero scalar part: $\pi_0(b) = \pi_0(a) - \pi_0(\mathbf{1}) = 1 - 1 = 0$. Define the algebraic inverse on group-like elements by the Neumann series in $b = a - \mathbf{1}$: \begin{align*} \psi : T_1((V)) &\to T((V)), \\ a &\mapsto \sum_{k = 0}^\infty (-b)^{\otimes k} = \sum_{k = 0}^\infty (\mathbf{1} - a)^{\otimes k}, \qquad b = a - \mathbf{1}. \end{align*} The two displayed forms are equal because $-b = \mathbf{1} - a$. The series converges in the product topology on $T((V))$: since $b$ has zero scalar part, the tensor power $b^{\otimes k}$ has $\pi_n(b^{\otimes k}) = 0$ for all $n < k$, so on each fixed level $n$ only the terms $k = 0, 1, \ldots, n$ contribute and the sum is finite. By the [reversal-inversion identity for signatures](/theorems/???), $\psi(S(x)) = S(\overleftarrow{x})$ for every $x \in \mathcal{C}_p$, so on $\mathcal{S}_p$ inversion factors as \begin{align*} [\,\cdot\,]^{-1} = S^{-1} \circ \psi|_{\mathcal{S}_p} \circ S. \end{align*}[/step]

[guided]We need to prove $[\,\cdot\,]^{-1}$ is continuous. Working directly with path classes is awkward; the standard manoeuvre is to lift everything through the signature map $S$, which makes the topology concrete (a subspace of the product of finite-dimensional spaces $V^{\otimes n}$). The factoring is \begin{align*} [\,\cdot\,]^{-1} = S^{-1} \circ \psi \circ S, \end{align*} where $\psi: T_1((V)) \to T((V))$ is the inversion at the level of the tensor algebra. **The Neumann series.** For $a \in T_1((V))$ — by definition, $\pi_0(a) = 1$ — set $b := a - \mathbf{1}$, so $a = \mathbf{1} + b$ and $\pi_0(b) = 0$. The standard formal manipulation \begin{align*} (\mathbf{1} + b)^{-1} = \sum_{k=0}^\infty (-b)^{\otimes k} \end{align*} suggests that the inverse of $a$ is the geometric series $\psi(a) := \sum_{k \ge 0} (-b)^{\otimes k} = \sum_{k \ge 0} (\mathbf{1} - a)^{\otimes k}$ (these are the same because $-b = -(a - \mathbf{1}) = \mathbf{1} - a$). **Why does the series converge?** In the product topology on $T((V))$, convergence means convergence of each level component. Fix $n \ge 0$ and consider $\pi_n((-b)^{\otimes k}) = (-1)^k \pi_n(b^{\otimes k})$. The crucial observation is that **$\pi_n(b^{\otimes k}) = 0$ whenever $k > n$**. To see this, expand $b^{\otimes k}$ using the convolution formula for the tensor product: \begin{align*} \pi_n(b^{\otimes k}) = \sum_{i_1 + \cdots + i_k = n} \pi_{i_1}(b) \otimes \pi_{i_2}(b) \otimes \cdots \otimes \pi_{i_k}(b), \end{align*} with each $i_j \ge 0$. Because $\pi_0(b) = 0$, every summand with any $i_j = 0$ vanishes, so we may restrict to compositions with all $i_j \ge 1$. But if $k > n$ and all $i_j \ge 1$, then $i_1 + \cdots + i_k \ge k > n$, contradicting the constraint $i_1 + \cdots + i_k = n$. Hence the index set is empty and $\pi_n(b^{\otimes k}) = 0$ for $k > n$. Consequently the level-$n$ component of the series truncates to a **finite** sum: \begin{align*} \pi_n(\psi(a)) = \sum_{k=0}^{n} (-1)^k \pi_n(b^{\otimes k}). \end{align*} Each level converges in $V^{\otimes n}$ as a finite sum, so the full series converges in the product topology on $T((V))$. **The output is in $T_1((V))$.** At level $0$: only $k = 0$ contributes (since $\pi_0(b^{\otimes k}) = 0$ for $k \ge 1$), giving $\pi_0(\psi(a)) = \pi_0(\mathbf{1}) = 1$. So $\psi(a) \in T_1((V))$. **Connection to inversion of paths.** By the [reversal-inversion identity](/theorems/???) for signatures, $S(\overleftarrow{x}) = S(x)^{-1}$ where the inverse is the Neumann series we just defined. So \begin{align*} [\,\cdot\,]^{-1} = S^{-1} \circ \psi|_{\mathcal{S}_p} \circ S \end{align*} factors inversion of path classes through inversion in $T_1((V))$. Once $\psi$ is shown continuous, $S$ is continuous, and $S^{-1}$ is continuous on its image, the composition $S^{-1} \circ \psi \circ S$ is continuous, which is exactly inversion on $\mathcal{C}_p$.[/guided]

[step:Verify $\psi(a)$ is a two-sided inverse of $a$ by direct level-wise computation]We verify that $\pi_n(a \otimes \psi(a)) = \pi_n(\mathbf{1})$ for all $n \ge 0$, i.e. $a \otimes \psi(a) = \mathbf{1}$ in $T((V))$. Write $a = \mathbf{1} + b$ with $b = a - \mathbf{1}$, $\pi_0(b) = 0$. Then $\psi(a) = \sum_{k \ge 0} (-b)^{\otimes k}$, and using the level-$n$ truncation, \begin{align*} a \otimes \psi(a) = (\mathbf{1} + b) \otimes \sum_{k \ge 0} (-b)^{\otimes k}. \end{align*} Project to level $n$ and split the sum at level $n$ on each side. For each fixed $n \ge 0$, the level-$n$ component of $a \otimes \psi(a)$ is determined by the components of $a$ and $\psi(a)$ up to level $n$, and using $\pi_n(b^{\otimes k}) = 0$ for $k > n$ from Step 2, \begin{align*} \pi_n(a \otimes \psi(a)) = \sum_{k=0}^{n} \pi_n\bigl((\mathbf{1} + b) \otimes (-b)^{\otimes k}\bigr) = \sum_{k=0}^{n} \pi_n\bigl((-b)^{\otimes k}\bigr) + \sum_{k=0}^{n} \pi_n\bigl(b \otimes (-b)^{\otimes k}\bigr). \end{align*} In the second sum, $b \otimes (-b)^{\otimes k} = -(-b) \otimes (-b)^{\otimes k} = -(-b)^{\otimes (k+1)}$. Substituting and re-indexing with $\ell = k + 1$, \begin{align*} \sum_{k=0}^{n} \pi_n\bigl(b \otimes (-b)^{\otimes k}\bigr) = \sum_{k=0}^{n} \pi_n\bigl(-(-b)^{\otimes (k+1)}\bigr) = -\sum_{\ell=1}^{n+1} \pi_n\bigl((-b)^{\otimes \ell}\bigr) = -\sum_{\ell=1}^{n} \pi_n\bigl((-b)^{\otimes \ell}\bigr), \end{align*} where in the last equality we used $\pi_n((-b)^{\otimes(n+1)}) = 0$ from Step 2 (the level-$n$ projection vanishes for tensor power $> n$). Combining, \begin{align*} \pi_n(a \otimes \psi(a)) = \sum_{k=0}^{n} \pi_n\bigl((-b)^{\otimes k}\bigr) - \sum_{\ell=1}^{n} \pi_n\bigl((-b)^{\otimes \ell}\bigr) = \pi_n\bigl((-b)^{\otimes 0}\bigr) = \pi_n(\mathbf{1}). \end{align*} The middle equality is the telescoping cancellation: the $k = 0$ term survives, every $k \ge 1$ term in the first sum is cancelled by the corresponding $\ell = k$ term in the second sum. Hence $a \otimes \psi(a) = \mathbf{1}$. The same computation with the order swapped gives $\psi(a) \otimes a = \mathbf{1}$, so $\psi(a) = a^{-1}$ in $T((V))$.[/step]

[guided]We promised a Neumann-series formula for the inverse but have not yet verified it actually inverts $a$. The verification is a direct level-wise telescoping computation — no formal manipulation that requires the series to "make sense as a whole". Fix $n \ge 0$. We compute $\pi_n(a \otimes \psi(a))$ explicitly and check it equals $\pi_n(\mathbf{1})$. Write $a = \mathbf{1} + b$ with $b = a - \mathbf{1}$. Since $\pi_0(b) = 0$, all the truncation lemmas from Step 2 apply: $\pi_m(b^{\otimes k}) = 0$ whenever $k > m$. **Decomposing $a \otimes \psi(a)$.** Using $a = \mathbf{1} + b$ and bilinearity of $\otimes$, \begin{align*} a \otimes \psi(a) = \psi(a) + b \otimes \psi(a). \end{align*} Project to level $n$. Using $\psi(a) = \sum_{k \ge 0}(-b)^{\otimes k}$ and the truncation $\pi_n((-b)^{\otimes k}) = 0$ for $k > n$, \begin{align*} \pi_n(\psi(a)) = \sum_{k=0}^{n} \pi_n\bigl((-b)^{\otimes k}\bigr). \end{align*} For the other term, we use $\pi_n(b \otimes \psi(a)) = \sum_{k \ge 0} \pi_n(b \otimes (-b)^{\otimes k})$. Now $b \otimes (-b)^{\otimes k} = -((-b) \otimes (-b)^{\otimes k}) = -(-b)^{\otimes (k+1)}$. So \begin{align*} \pi_n(b \otimes \psi(a)) = -\sum_{k \ge 0} \pi_n\bigl((-b)^{\otimes (k+1)}\bigr) = -\sum_{\ell \ge 1} \pi_n\bigl((-b)^{\otimes \ell}\bigr) = -\sum_{\ell=1}^{n} \pi_n\bigl((-b)^{\otimes \ell}\bigr), \end{align*} where we re-indexed $\ell = k + 1$ and truncated using $\pi_n((-b)^{\otimes \ell}) = 0$ for $\ell > n$. **Telescope.** Adding, \begin{align*} \pi_n(a \otimes \psi(a)) = \pi_n(\psi(a)) + \pi_n(b \otimes \psi(a)) = \sum_{k=0}^{n} \pi_n\bigl((-b)^{\otimes k}\bigr) - \sum_{\ell=1}^{n} \pi_n\bigl((-b)^{\otimes \ell}\bigr). \end{align*} The $k = 0$ term in the first sum is not cancelled; every $k \ge 1$ term is cancelled by its $\ell = k$ counterpart in the second sum. The result is \begin{align*} \pi_n(a \otimes \psi(a)) = \pi_n\bigl((-b)^{\otimes 0}\bigr) = \pi_n(\mathbf{1}). \end{align*} Since this holds for every level $n$, $a \otimes \psi(a) = \mathbf{1}$ in $T((V))$. The mirror computation $\psi(a) \otimes a = \mathbf{1}$ proceeds identically — replace each $b \otimes (-b)^{\otimes k}$ by $(-b)^{\otimes k} \otimes b$ and observe that the sign-flip $(-b)^{\otimes k} \otimes b = -(-b)^{\otimes(k+1)}$ yields the same telescope. So $\psi(a)$ is the two-sided inverse of $a$ in $T((V))$ (and lies in $T_1((V))$ as already shown), justifying the formula. **Why the careful telescoping matters.** The naive "Neumann series cancellation" $a (\mathbf{1} - (\mathbf{1} - a) + (\mathbf{1} - a)^2 - \cdots) = \mathbf{1}$ is a formal identity that requires the series to multiply distributively without convergence concerns. In a topological algebra, we must check that the level-$n$ projection of the cancellation actually terminates — which it does, exactly because $b$ has zero scalar part. Without that hypothesis (e.g. if $\pi_0(a) \ne 1$), the series would not truncate level by level and the Neumann formula would fail.[/guided]

[step:Reduce continuity of $\psi$ to continuity of each $\pi_n \circ \psi$]By the [universal property of the product topology](/theorems/???), a map $\psi: T_1((V)) \to T((V))$ is continuous if and only if $\pi_n \circ \psi$ is continuous for each $n \ge 0$. Fix $n \ge 0$. From Step 2, only the terms $k = 0, 1, \ldots, n$ contribute to $\pi_n(\psi(a))$, so \begin{align*} \pi_n \circ \psi(a) = \sum_{k=0}^{n} (-1)^k \pi_n\bigl(b^{\otimes k}\bigr), \qquad b = a - \mathbf{1}. \end{align*} This expression depends only on $\pi_0(a), \pi_1(a), \ldots, \pi_n(a)$, since $\pi_n(b^{\otimes k})$ involves only the components of $b$ — equivalently, of $a$ — up to level $n$.[/step]

[guided]By the universal property, continuity into a product reduces to continuity into each factor: $\psi$ is continuous if and only if $\pi_n \circ \psi: T_1((V)) \to V^{\otimes n}$ is continuous for every $n$. We work with one $n$ at a time. For fixed $n$, the truncation result from Step 2 gives the finite formula \begin{align*} \pi_n \circ \psi(a) = \sum_{k=0}^{n} (-1)^k \pi_n\bigl(b^{\otimes k}\bigr), \end{align*} where $b = a - \mathbf{1}$. Each $\pi_n(b^{\otimes k})$ is a finite multilinear expression in the level components $\pi_0(b), \ldots, \pi_n(b)$ (and, since $\pi_0(b) = 0$ on $T_1((V))$, only the levels $\pi_1(b), \ldots, \pi_n(b)$ actually contribute). Linear in $a$: $\pi_j(b) = \pi_j(a) - \pi_j(\mathbf{1})$, so the components of $b$ are continuous functions of the components of $a$. Hence $\pi_n \circ \psi$ depends continuously on the finite tuple $(\pi_0(a), \ldots, \pi_n(a))$, and the next step shows that "depends continuously" really does mean "is continuous".[/guided]

[step:Establish continuity of $\pi_n \circ \psi$ via multilinearity in finitely many components]We show each summand $a \mapsto \pi_n((-b)^{\otimes k})$ is continuous on $T_1((V))$ (with $b = a - \mathbf{1}$ and $0 \le k \le n$); the finite sum then gives continuity of $\pi_n \circ \psi$. The map $a \mapsto b = a - \mathbf{1}$ is continuous (translation in $T((V))$ preserves the product topology). Project to levels: $a \mapsto (\pi_1(b), \pi_2(b), \ldots, \pi_n(b)) \in \prod_{j=1}^n V^{\otimes j}$ is continuous by definition of the product topology (recall $\pi_0(b) = 0$ on $T_1((V))$, so the level-$0$ slot contributes nothing). For the level-$n$ projection of $b^{\otimes k}$, expand using the convolution structure: \begin{align*} \pi_n(b^{\otimes k}) = \sum_{(i_1, \ldots, i_k)} \pi_{i_1}(b) \otimes \pi_{i_2}(b) \otimes \cdots \otimes \pi_{i_k}(b), \end{align*} where the sum is over compositions $(i_1, \ldots, i_k)$ of $n$ with each $i_j \ge 1$ (compositions with $i_j = 0$ vanish since $\pi_0(b) = 0$). This is a finite sum (the number of compositions of $n$ into $k$ positive parts is $\binom{n-1}{k-1}$, and we need $k \le n$). Each summand is a $k$-linear map \begin{align*} V^{\otimes i_1} \times V^{\otimes i_2} \times \cdots \times V^{\otimes i_k} &\to V^{\otimes n}, \\ (\beta_1, \ldots, \beta_k) &\mapsto \beta_1 \otimes \cdots \otimes \beta_k, \end{align*} between finite-dimensional vector spaces; multilinear maps between finite-dimensional vector spaces are continuous (immediate from equivalence of norms in finite dimensions). The composition with the continuous projection $a \mapsto (\pi_{i_1}(b), \ldots, \pi_{i_k}(b))$ yields continuity of $a \mapsto \pi_n(b^{\otimes k})$. Multiplying by $(-1)^k$ preserves continuity. Summing over $k = 0, 1, \ldots, n$ — a finite sum — preserves continuity. Hence $\pi_n \circ \psi$ is continuous, and by Step 4 so is $\psi$. By Step 2, $\psi: \mathcal{S}_p \to \mathcal{S}_p$ (since $\psi(S(x)) = S(\overleftarrow{x}) \in \mathcal{S}_p$), and $\psi$ is continuous on $T_1((V))$ hence continuous on $\mathcal{S}_p$ by restriction. The composition $S^{-1} \circ \psi \circ S$ is therefore continuous on $\mathcal{C}_p$, proving inversion is continuous.[/step]

[guided]We have reduced to showing that for each $n$, the map $\pi_n \circ \psi: T_1((V)) \to V^{\otimes n}$ is continuous. The structure is: \begin{align*} T_1((V)) \xrightarrow{\;a \mapsto (\pi_1(b), \ldots, \pi_n(b))\;} \prod_{j=1}^{n} V^{\otimes j} \xrightarrow{\;F_n\;} V^{\otimes n}, \end{align*} where $b = a - \mathbf{1}$, the first arrow is continuous by the product-topology definition, and $F_n$ is an explicit polynomial in tensor coordinates. **What is $F_n$?** Substituting $\pi_n(b^{\otimes k}) = \sum \pi_{i_1}(b) \otimes \cdots \otimes \pi_{i_k}(b)$ (sum over compositions of $n$ into $k$ positive parts) into Step 4's formula, \begin{align*} F_n(\beta_1, \ldots, \beta_n) := \sum_{k=0}^{n} (-1)^k \sum_{\substack{i_1 + \cdots + i_k = n \\ i_j \ge 1}} \beta_{i_1} \otimes \beta_{i_2} \otimes \cdots \otimes \beta_{i_k}, \end{align*} where $\beta_j$ stands for the level-$j$ component (the $k = 0$ term contributes $\pi_n(\mathbf{1})$, which is $0$ unless $n = 0$, in which case it is $1 \in V^{\otimes 0} = \mathbb{R}$). Each interior sum is a finite collection of $k$-linear maps from $V^{\otimes i_1} \times \cdots \times V^{\otimes i_k}$ to $V^{\otimes n}$. **Why are these multilinear maps continuous?** Multilinear maps on finite-dimensional vector spaces are continuous. To see this, fix bases of each $V^{\otimes i_j}$ and write the multilinear map as a polynomial in the coordinates with values in coordinates of $V^{\otimes n}$; polynomials in finitely many real variables are continuous. (Equivalently: any norm makes each $V^{\otimes i_j}$ a finite-dimensional normed space, and multilinear maps are bounded with respect to the product norm, hence continuous.) **Why is the sign tracking unambiguous?** Each summand is $(-b)^{\otimes k}$ projected to level $n$, which expands as $(-1)^k$ times the convolution sum of level components of $b$. The factor $(-1)^k$ is uniform across the entire $k$-th sum — it depends only on the tensor-power index $k$, not on the individual composition $(i_1, \ldots, i_k)$. Compositions with any $i_j = 0$ are killed by $\pi_0(b) = 0$, so we restrict to compositions with all $i_j \ge 1$, which simply removes those terms; it does not change the sign of the surviving ones. **Total $F_n$ is continuous.** $F_n$ is a finite sum (over $k = 0, 1, \ldots, n$ and over compositions of $n$ into $k$ positive parts, finitely many in total) of constant-multiplied multilinear maps between finite-dimensional spaces, hence continuous. By the universal-property argument from Step 4, $\psi$ itself is continuous. The composition $S^{-1} \circ \psi \circ S$ is continuous, so inversion on $\mathcal{C}_p$ is continuous.[/guided]

[step:Establish continuity of multiplication by the parallel argument]Define \begin{align*} \mu : T_1((V)) \times T_1((V)) &\to T_1((V)), \\ (a, b) &\mapsto a \otimes b. \end{align*} By Chen's identity, the multiplication on $\mathcal{C}_p$ factors as \begin{align*} [\,\cdot\,] \cdot [\,\cdot\,] = S^{-1} \circ \mu|_{\mathcal{S}_p \times \mathcal{S}_p} \circ (S \times S). \end{align*} $S \times S$ is continuous (product of continuous maps) and $S^{-1}$ is continuous on $\mathcal{S}_p$. It remains to show $\mu$ is continuous. By the universal property of the product topology on the codomain $T((V))$, $\mu$ is continuous iff $\pi_n \circ \mu$ is continuous for every $n \ge 0$. The graded product gives \begin{align*} \pi_n \circ \mu(a, b) = \pi_n(a \otimes b) = \sum_{i + j = n} \pi_i(a) \otimes \pi_j(b), \end{align*} a finite sum (over $0 \le i, j \le n$ with $i + j = n$) of bilinear maps $V^{\otimes i} \times V^{\otimes j} \to V^{\otimes n}$. Each such bilinear map between finite-dimensional spaces is continuous, and the sum depends only on the finitely many continuous coordinate projections $\pi_0(a), \ldots, \pi_n(a)$ and $\pi_0(b), \ldots, \pi_n(b)$. Hence $\pi_n \circ \mu$ is continuous, and $\mu$ is continuous. Therefore multiplication on $\mathcal{C}_p$ is continuous. Combined with continuity of inversion, $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ is a topological group.[/step]

[guided]The argument for multiplication is structurally identical to the argument for inversion — the only difference is that multiplication is *bilinear* rather than (in the relevant sense) multilinear-in-one-argument. The level-$n$ piece of $a \otimes b$ is determined by the formula \begin{align*} \pi_n(a \otimes b) = \sum_{i + j = n} \pi_i(a) \otimes \pi_j(b), \end{align*} which involves only finitely many tensor levels of $a$ and $b$ (namely $i, j \le n$). For each pair $(i, j)$ with $i + j = n$, the bilinear pairing \begin{align*} V^{\otimes i} \times V^{\otimes j} \to V^{\otimes n}, \qquad (\alpha, \beta) \mapsto \alpha \otimes \beta \end{align*} is continuous because it is a bilinear map between finite-dimensional real vector spaces (and bilinear maps in finite dimensions are always continuous — this is one of the standard consequences of equivalence of norms in finite dimensions). The finite sum of continuous bilinear maps is continuous. So $\pi_n \circ \mu$ is continuous; by the universal property, $\mu$ is continuous; hence multiplication on $\mathcal{C}_p$ is continuous. Both group operations being continuous, $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ is a topological group.[/guided]