Forgetting the last $n$ points is locally trivial with fibre $F_n$ of the punctured manifold

Chart data. By [L2] and [L7] there are charts ${\varphi }_j:V_j\to O_j$ with $q_j\in V_j$, $O_j\subseteq {\mathbb{R}}^d$ open, ${\varphi }_j(q_j)=0$, and the $V_j$ pairwise disjoint (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not); $m$ choices are made and no infinite selection occurs. Shrinking $V_j$ if necessary to the inverse image of an open ball, we may suppose $\bar{B}(0,4{\rho }_j)\subseteq O_j$ for some ${\rho }_j>0$. Put $S_j:={\varphi }_j^{-1}(B(0,2{\rho }_j))\subseteq V_j$ and $W_j:={\varphi }_j^{-1}(B(0,{\rho }_j))\subseteq S_j$, and finally $U:=\{x\in F_m(M):x_j\in W_j\text{ for all }j\}$, which is $F_m(M)\cap (W_1\times \cdots \times W_m)$ and hence open in $F_m(M)$ with $q\in U$.

The bump function. Fix $j$ and put $b(z):=\max (0,1-\parallel z\parallel \mathrm{/}(2{\rho }_j))$ for $z\in {\mathbb{R}}^d$. Then $0\le b\le 1$, $b(0)=1$, $b(z)=0$ for $\parallel z\parallel \ge 2{\rho }_j$, $b$ is continuous by [L5] and [L8], and $b$ is Lipschitz with constant $1\mathrm{/}(2{\rho }_j)$: for $z,z^{\mathrm{\prime }}\in {\mathbb{R}}^d$ one has $\mathrm{\mid }b(z)-b(z^{\mathrm{\prime }})\mathrm{\mid }\le \mid \parallel z\parallel -\parallel z^{\mathrm{\prime }}\parallel \mid \mathrm{/}(2{\rho }_j)\le \parallel z-z^{\mathrm{\prime }}\parallel \mathrm{/}(2{\rho }_j)$ by [L3].

The radial mover of the coordinate space. Fix $j$, let $b$ be as in step 1.2 and let $u\in {\mathbb{R}}^d$ with $\parallel u\parallel \le {\rho }_j$; put ${\theta }_u(z):=z+b(z)u$. Then: ${\theta }_u$ is continuous with $\parallel {\theta }_u(z)-z\parallel \le \parallel u\parallel$ and ${\theta }_u(z)=z$ whenever $\parallel z\parallel \ge 2{\rho }_j$; ${\theta }_u$ is injective, since $\parallel {\theta }_u(z)-{\theta }_u(z^{\mathrm{\prime }})\parallel \ge \parallel z-z^{\mathrm{\prime }}\parallel -\mathrm{\mid }b(z)-b(z^{\mathrm{\prime }})\mathrm{\mid }\parallel u\parallel \ge (1-\frac{\parallel u\parallel }{2{\rho }_j})\parallel z-z^{\mathrm{\prime }}\parallel \ge \frac{1}{2}\parallel z-z^{\mathrm{\prime }}\parallel$; and ${\theta }_u$ is surjective, because for $w\in {\mathbb{R}}^d$ the map $T(y):=w-b(y)u$ satisfies $\parallel T(y)-T(y^{\mathrm{\prime }})\parallel \le \frac{\parallel u\parallel }{2{\rho }_j}\parallel y-y^{\mathrm{\prime }}\parallel \le \frac{1}{2}\parallel y-y^{\mathrm{\prime }}\parallel$ and ${\mathbb{R}}^d$ is complete, so by [L4] it has a fixed point $y=T(y)$, which says exactly ${\theta }_u(y)=w$. Hence ${\theta }_u$ is a bijection of ${\mathbb{R}}^d$ fixing the complement of $B(0,2{\rho }_j)$, and ${\theta }_u(0)=u$.

The inverse family and its Lipschitz estimate. With the notation of step 2.1, let $y:={\theta }_u^{-1}(w)$ and $y^{\mathrm{\prime }}:={\theta }_{u^{\mathrm{\prime }}}^{-1}(w^{\mathrm{\prime }})$ for $\parallel u\parallel ,\parallel u^{\mathrm{\prime }}\parallel \le {\rho }_j$. Since $y=w-b(y)u$ and $y^{\mathrm{\prime }}=w^{\mathrm{\prime }}-b(y^{\mathrm{\prime }})u^{\mathrm{\prime }}$, the triangle inequality and the Lipschitz bound of step 1.2 give $\parallel y-y^{\mathrm{\prime }}\parallel \le \parallel w-w^{\mathrm{\prime }}\parallel +\frac{\parallel u\parallel }{2{\rho }_j}\parallel y-y^{\mathrm{\prime }}\parallel +\parallel u-u^{\mathrm{\prime }}\parallel \le \parallel w-w^{\mathrm{\prime }}\parallel +\frac{1}{2}\parallel y-y^{\mathrm{\prime }}\parallel +\parallel u-u^{\mathrm{\prime }}\parallel ,$ hence $\parallel {\theta }_u^{-1}(w)-{\theta }_{u^{\mathrm{\prime }}}^{-1}(w^{\mathrm{\prime }})\parallel \le 2(\parallel u-u^{\mathrm{\prime }}\parallel +\parallel w-w^{\mathrm{\prime }}\parallel )$. In particular each ${\theta }_u^{-1}$ is continuous, so ${\theta }_u$ is a homeomorphism of ${\mathbb{R}}^d$ by step 2.1, and $(u,w)\mapsto {\theta }_u^{-1}(w)$ is continuous on $\{u:\parallel u\parallel \le {\rho }_j\}\times {\mathbb{R}}^d$. Moreover ${\theta }_u^{-1}(w)=w-b({\theta }_u^{-1}(w))u$, so $\parallel {\theta }_u^{-1}(w)-w\parallel \le \parallel u\parallel \le {\rho }_j$, and ${\theta }_u^{-1}(w)=w$ whenever $\parallel w\parallel \ge 2{\rho }_j$; consequently ${\theta }_u^{-1}$ maps $B(0,3{\rho }_j)$ into $B(0,4{\rho }_j)\subseteq O_j$.

Point-moving homeomorphisms of $M$. Fix $j$ and $x\in U$, and put $u_j:={\varphi }_j(x_j)$. Since ${\theta }_{j,u_j}$ is a bijection fixing the complement of $B(0,2{\rho }_j)$ pointwise, it carries that ball onto itself; its inverse has the same property. Set $K_j:={\varphi }_j^{-1}(\bar{B}(0,2{\rho }_j))$. By [L9], $K_j$ is compact and closed in $M$, and $K_j\subset V_j$. On the open cover $V_j,M\setminus K_j$ define $h_{j,x}$ by ${\varphi }_j^{-1}{\theta }_{j,u_j}{\varphi }_j$ on $V_j$ and by the identity on $M\setminus K_j$. The chart formula is defined on all of $V_j$: it preserves the ball and fixes every point of $O_j$ outside it. The two formulas agree on $V_j\setminus K_j$, where ${\theta }_{j,u_j}$ is the identity, so [L5] gives continuity. Replacing ${\theta }_{j,u_j}$ by its inverse gives a continuous map $h_{j,x}^{\mathrm{\prime }}$ on the same cover. Both maps preserve $V_j$, their chart formulas are mutually inverse, and outside $V_j$ both are the identity; hence they are inverse homeomorphisms of $M$. Moreover $h_{j,x}(q_j)=x_j$, the map fixes $M\setminus S_j$ pointwise, and it fixes $q_k$ for $k\mathrm{\ne }j$.

Joint continuity of the point-moving family. On $U\times V_j$ the formula $(x,y)\mapsto {\varphi }_j^{-1}({\theta }_{j,{\varphi }_j(x_j)}({\varphi }_j(y)))$ is jointly continuous by the continuity of the chart, coordinate projections, and the vector operations in ${\theta }_{j,u}(z)=z+b_j(z)u$. On $U\times (M\setminus K_j)$ the formula is $(x,y)\mapsto y$. These open sets cover $U\times M$ and the formulas agree on their overlap by step 4.1. Thus [L5] proves joint continuity of $(x,y)\mapsto h_{j,x}(y)$. The inverse family is jointly continuous by the identical open-cover argument using the estimate of step 3.1.

The family $h_x$ and its inverse. For $x\in U$ put $h_x:=h_{1,x}\circ h_{2,x}\circ \cdots \circ h_{m,x}$ and $h_x^{-1}:=h_{m,x}^{\mathrm{\prime }}\circ \cdots \circ h_{2,x}^{\mathrm{\prime }}\circ h_{1,x}^{\mathrm{\prime }}$. Each factor is a homeomorphism of $M$ supported in the pairwise disjoint open sets $V_j$, so the factors commute and the two displayed composites are inverse to each other; hence $h_x$ is a homeomorphism of $M$ for every $x\in U$. Moreover $h_x(q_j)=x_j$ for every $j$, because every factor with index $k\mathrm{\ne }j$ fixes $q_j\in V_j\subseteq M\setminus V_k$ by step 4.1. By step 5.1 and [L5] the maps $(x,y)\mapsto h_x(y)$ and $(x,y)\mapsto h_x^{-1}(y)$ are continuous on $U\times M$. Since $h_x$ carries the finite set $Q$ bijectively onto $\{x_1,\dots ,x_m\}$, it restricts to a bijection $M\setminus Q\to M\setminus \{x_1,\dots ,x_m\}$, and hence induces a bijection $F_n(M\setminus Q)\to F_n(M\setminus \{x_1,\dots ,x_m\})$ by acting on coordinates.

The trivialization is well defined. Define $\mathrm{\Phi }(x,y):=(x_1,\dots ,x_m,h_x(y_1),\dots ,h_x(y_n))$ for $x\in U$ and $y\in F_n(M\setminus Q)$. The $m+n$ displayed points are pairwise distinct: the $x_j$ are pairwise distinct and so are the $h_x(y_k)$ by step 6.1, while $h_x(y_k)\mathrm{\ne }{x}_j=h_x(q_j)$ because $y_k\mathrm{\ne }{q}_j$ for $y\in F_n(M\setminus Q)$. Hence $\mathrm{\Phi }$ takes values in $F_{m+n}(M)$, and $\pi (\mathrm{\Phi }(x,y))=x$ by construction, so $\mathrm{\Phi }$ maps $U\times F_n(M\setminus Q)$ into ${\pi }^{-1}(U)\subseteq F_{m+n}(M)$ and $\pi \circ \mathrm{\Phi }={\mathrm{pr}}_1$ holds there.

$\mathrm{\Phi }$ is continuous. The first $m$ components of $\mathrm{\Phi }$ are the projections of $x$, which are continuous by [L6]; the $k$-th forgotten coordinate is $(x,y)\mapsto h_x(y_k)$, the composite of the continuous map $(x,y)\mapsto (x,y_k)$ with the jointly continuous map $(x,y)\mapsto h_x(y)$ of step 6.1; the domain is the subspace $U\times F_n(M\setminus Q)\subseteq U\times M^n$, and restrictions of continuous maps are continuous. Hence $\mathrm{\Phi }$ is continuous as a map into the subspace $F_{m+n}(M)$ of $M^{m+n}$ by [L6].

The inverse trivialization. For $(x,z)\in {\pi }^{-1}(U)$, that is $x\in U$ and $z\in F_n(M\setminus \{x_1,\dots ,x_m\})$, put $\mathrm{\Psi }(x,z):=(x,h_x^{-1}(z_1),\dots ,h_x^{-1}(z_n))$, where the first component is the base point $x$ and the second the $n$-tuple of inverse images. Since $h_x^{-1}$ is injective and $z_k\mathrm{\ne }{x}_j=h_x(q_j)$ for all $k,j$, the points $h_x^{-1}(z_k)$ are pairwise distinct and all outside $Q$, so $\mathrm{\Psi }$ takes values in $U\times F_n(M\setminus Q)$; it is continuous by step 6.1 and [L6] exactly as in step 7.2, and $\mathrm{\Psi }(\mathrm{\Phi }(x,y))=(x,y)$, $\mathrm{\Phi }(\mathrm{\Psi }(x,z))=(x,z)$ because $h_x^{-1}$ inverts $h_x$.

Conclusion. By steps 7.1, 7.2 and 8.1 the map $\mathrm{\Phi }:U\times F_n(M\setminus Q)\to {\pi }^{-1}(U)$ is a continuous bijection with continuous inverse, hence a homeomorphism, and $\pi \circ \mathrm{\Phi }={\mathrm{pr}}_1$; restricting $\mathrm{\Phi }$ to $\{x\}\times F_n(M\setminus Q)$ exhibits the fibre ${\pi }^{-1}(x)$ over any $x\in U$ as homeomorphic to $F_n(M\setminus Q)$. This is precisely a local trivialization of $\pi$ at the base configuration $q$ with fibre $F_n(M\setminus Q)$, and since $q$ was arbitrary the map is locally trivial at every base configuration. The construction used only finitely many choices of charts and radii, so no choice principle is used.