Continuity of a potential $(n+1$)-simplex from $n$-simplex: Rotman "Introduction to Algebraic Topology" This is from Rotman's book Introduction to Algebraic Topology (p.72-73).
He starts with a convex subset $X$ of a Euclidean space, an $n$-simplex $\sigma\colon \Delta^n\to X$ and a point $b$ of $X$. He then defines $(n+1)$-simplex $b.\sigma\colon \Delta^{n+1}\to X$ as follows: $(b.\sigma)(x_0,x_1,\dots,x_n)$ is equal to $b$ if $(x_0,x_1,\dots,x_n) = (1,0,\dots,0)$ and to $x_0b + (1-x_0)\sigma\left(\dfrac{x_1}{1-x_0}, \dots, \dfrac{x_n}{1-x_0}\right)$ otherwise.
Rotman explains why this map is well-defined, and I can follow this explanation just fine. The problem lies in seeing why it is continuous.

*

*The first idea is to use so-called "gluing lemma": given a map from a topological space which is a union of two open or two closed open subsets which agree on their intersection, we can check it's continuity by checking the continuity of respective restrictions. However, it doesn't seem to apply here: $\{(1,0,\dots,0)\}$ is closed in $\Delta^{n+1}$ since the latter is Hausdorff, so $\Delta^{n+1}\setminus\{(1,0,\dots,0)\}$ is open. Since $\Delta^{n+1}$ is path-connected, the only "clopen" subsets are $\varnothing$ and $\Delta^{n+1}$ itself.

*The continuity of $b.\sigma$ at $(x_0,x_1,\dots,x_n) \neq (1,0,\dots,0)$ doesn't seem that hard at first. Indeed, the map $(x_0,x_1,\dots,x_n) \mapsto \left(\dfrac{x_1}{1-x_0}, \dots, \dfrac{x_n}{1-x_0}\right)$ is continuous as it is continuous coordinate-wise, $\sigma$ is continuous by assumption, and a multiplication and a sum of two continuous map is continuous. However, it would be two if the entire map $b.\sigma$ could be described like that, but it is defined point-wise. And, as I've said earlier, we can't separate the map into restrictions here.

*Also, I don't see how to check continuity of $b.\sigma$ at $(1,0,\dots,0)$.

I have to confess that I don't know much of the theory of simplices, so I would prefer as elementary solution as possible. However, if some of it is needed, I would be extra grateful for a reference for the facts used.
 A: As you explained, the continuity on the open subset $\Delta^{n+1} \setminus
 \{(1,0\ldots,0)\}$ is easily verified. It remains to show continuity at $e_1 = (1,0,\ldots,0)$. The gluing lemma does not help us, we have to give a direct proof.
Write $x \in \Delta^{n+1}$ in the form $x = (x_0,\xi)$ with $x_0 \in \mathbb R$ and $\xi \in \mathbb R^n$. Then for $x \ne e_1$
$$(b.\sigma)(x) = x_0b + (1-x_0)\sigma(\frac{1}{1-x_0}\xi) .$$
Since $\Delta^n$ is compact, $\sigma(\Delta^n)$ is compact, thus bounded. That is, we find $R \ge 0$ such that $\lVert \sigma(y) \rVert \le R$ for all $y \in \Delta^n$. This shows that for $x \ne e_1$
$$\lVert (b.\sigma)(x) - e_1 \rVert = \lVert (x_0 -1) b + (1-x_0)\sigma(\frac{1}{1-x_0}\xi) \rVert = \lvert x_0 -1 \rvert \cdot \lVert  b - \sigma(\frac{1}{1-x_0}\xi) \rVert \\ \le \lvert x_0 -1 \rvert \cdot(\lVert  b \rVert + \lVert \sigma(\frac{1}{1-x_0}\xi) \rVert) \le \lvert x_0 -1 \rvert \cdot(\lVert  b \rVert + R) .$$
Sincbe $\lvert x_0 -1 \rvert \le \lVert x - e_1 \rVert$ we get
$$\lVert (b.\sigma)(x) - e_1 \rVert < (\lVert  b \rVert + R)\lVert x - e_1 \rVert $$
which proves continuity.
