Barycentric Coordinates are Continuous Functions While studying some basic homology stuff, I encountered the idea of Barycentric Coordinates. I am using Elements of Algebraic Topology by James Munkres. On p.g. 4 of his book, Munkres roughly states that barycentric coordinates are continuous functions ($$the\ barycentric\ coordinates\ t_i(x) \ of \ x $$ with respect to $$a_0,..a_n\ are\ continuous\ functions\ of\ x$$). I do not understand how this is working. Some enlightenment on this would be highly appreciated. Thank you.
 A: Notation and definitions will be that of Munkres' Elements of Algebraic Topology.
Throughout, let $x, y$ be points in the $n$-plane spanned by the geometrically independent set $a = \{a_0, \dots, a_n\} \subset \Bbb{R}^N$.
Property (1) on page 5 reads:

The barycentric coordinates $t_i(x)$ of $x$ with respect to $a$ are continuous functions of $x$.

It is already shown in an earlier exercise that indeed the values $t_i(x), i =0..n$ are uniquely determined by $x$, so that we indeed have functions of $x$.
Define the $n$-plane $P$ to be the set of points spanned by the vectors in $a$ or all $z = \sum_{i=0}^n \lambda_i a_i$ such that $\lambda_i \in \Bbb{R}$.  This is a superset of the simplex $\sigma$ which is defined by having the extra condition that $\sum_{i=0}^n t_i = 1$.
Thus it suffices to prove continuity of $t_i : P \to \Bbb{R}$ and then continuity on $\sigma$ follows via restriction of a continuous map which is then in turn continuous.

Which definition of continuity of a function $P \to \Bbb{R}$ should we use, where $P \subset \Bbb{R}^N$ is a vector subspace?
This is the difficult part to answer, since once we have a suitable definition to use, it's just a matter of expanding the definition.
First of all, are the $t_i(x)$ linear functions?  Since then we automatically get continuity.

We know that if $x - y = \sum_{i=0}^n (t_i(x) - t_i(y)) a_i$.  But such coefficients are uniquely determined by $x-y$ and called $t_i(x-y)$.  Thus we have $t_i(x-y) = t_i(x) - t_i(y)$.  That is equivalent to $t_i(x+y) = t_i(x) + t_i(y)$ for all $x,y \in \Bbb{R}^N$.
And $\alpha x = \sum_{i=0}^n \alpha t_i a_i$.   But we know that $\alpha x$ uniquely determines its $t_i$ so that $t_i(\alpha x) = \alpha t_i(x)$ for all $\alpha \in \Bbb{R}$.
Thus we've proven that $t_i$'s are linear maps of $x \in \Bbb{R}^N$.  Every linear map on a finite dimensional vector space is continuous.  See further answers on MSE for proof.
$\blacksquare$
A: This looks trivial and it is. But one might not be able to solve or it might cause confusion when one try proving by directly writing down. I would recommend other solution over direct proof. Some might be curious and get distracted during the proof. I guess for that case,
"Every linear mapping on a finite dimensional space is continuous"
is the topic you might want.
Direct proof is not recommended and might not be possible(I'm not sure though). Since we are only given geometric independency, we are not provided with information to write down distance with given ${a_i}$. Of course it would be possible when we have specific case, but generally, I am inclined to opinion that is not possible.
Note that as a function, $t_i(x) $is equivalent to composition of three functions. It would be projection $\circ$ isomorphism $\circ$ translation. Then besides trivial continuity, you would want to prove "isomorphism constucted by assigning basis of domain and codomain is indeed continuous" . For our, $\mathbb{R}^N$ case, fact that "all linear functions of finite dimensional Euclidean space is continuous" would suffice and you can check it's proof in analysis text book if it has multivariable part.(Cauchy Schwarz inequality is the key)
You might be curious and search for more general result at this point, then check the above link.
For our function is composition of contiuous function, it is continuous.
