I'm trying to solve the following exercise:
Let $X$ be a compact Hausdorff space and let $U_1,...,U_n$ be a cover of $X$ of order $m$. Let $z_1,...,z_n\in\mathbb{R}^N$ for some $N$ be in general position. Suppose $h_1,...,h_n:X\to[0,1]$ are a partition of unity subordinate to the given cover.
Define $F:X\to\mathbb{R}^N$ by $F(x)=\sum_{k=1}^nh_k(x)z_k$.
Show that $F(X)$ has covering dimension at most $m-1$.
What I tried so far: A first guess was to try to use the images of $U_1,...,U_n$ under $F$, but as we don't know that $F$ is open, these images aren't necessarily open.
Using the assumption that the given cover is of order $m$, we know that for any given $x\in X$, there are only $m$ summands in $F(x)$, and hence I looked at the following sets: for any $I\subseteq[1,n]$ such that $|I|=m$ denote $Y_I=X\setminus\left(\bigcup_{i\not\in I} U_i\right)$. Then $Y_I$ are closed and $F(X)=\bigcup_I \overline{F(Y_I)}$ and thus it's sufficient to prove that $\overline{F(Y_I)}$ has covering dimension at most $m-1$.
Note that for every such $I$ and every $y\in Y_I$, $F(y)$ has only $m$ summands at most.
But as I do not know the connection between $N$ and $m$, here I got stuck.
Any help will be greatly appreciated.