A continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}^m$, with $m > n$, Consider some continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}^m$, with $m > n$.  For notation, $\mathbf{Y} = g(\mathbf{X})$ where ${\bf X} \in \mathbb{R}^n$ and ${\bf Y} \in \mathbb{R}^m$.  We can say also write $Y_i = g_i(\mathbf{X})$ for each individual output, with $g_i$ being a continuous function. 
What conditions would we require to be able to express at least one of the output variables, say $Y_m$ as a function of the other outputs instead? Wlog say $Y_m =h(Y_1,...Y_{m-1})$, for some continuous function $h$.  
Let $Y_{\bf n} = (Y_1,...,Y_n) = (g_1({\bf X}),...,g_n({\bf X})) = g_{\bf n}({\bf X})$.  For example, if $g_{\bf n}$ is bijective then we know there is an inverse such that $\mathbf{X} = g_{\bf n}^{-1}({\bf Y}_{\bf n})$.  We could therefore say $Y_k = g_k(g_{\bf n}^{-1}({\bf Y}_{\bf n}))$,$\forall k > n$.  However it seems to me that $g_{\bf n}$ being bijective is quite restrictive.  It seems intuitive to me the fact that we are mapping into a larger space means our output be some manifold with a smaller dimensionality than $m$, but I'm not entirely sure how to see this, if it is true. 
edit:  Fixed some notation errors.
 A: The answer directly depends not on the map $g$, but on its image $Z=g(\Bbb R^n)$. Indeed, let $\pi_m:\Bbb R^m\to\Bbb R^{m-1}$ and $p_m:\Bbb R^m\to\Bbb R$ be projections, $\pi_m(Y_1,\dots,Y_m)=(Y_1, \dots,Y_{m-1})$ and $p_m(Y_1,\dots,Y_m)=Y_m$ for each $Y=(Y_1,\dots,Y_m)\in\Bbb R^m$. If there exists a function $h:\pi_m(Z)\to\Bbb R$ such that $Y_m=h\pi_m(Y)$ for each $Y\in Z$ then a restriction $\pi_m|Z$ is injective. Conversely, if  a map $\pi_m|Z$ is injective then we can put  $h(Y’)=p_m((\pi_m|Z)^{-1}(Y’))$ for each point $Y’\in\pi_m(Z)$.
Thus $h$ is a unique map from $\pi_m(Z)$ to $\Bbb R$ such that $p_m|Z=h\circ(\pi_m|Z)$. We claim that the map $h$ is continuous iff the map $\pi_m|Z$ is open onto image, that is $(\pi_m|Z)(V)$ is open in $\pi_m(Z)$ for each open subset $V$ of $Z$. Indeed, assume that the map $\pi_m|Z$ is open onto image. Let $U\subset R$ be any open set. Then a set 
$$h^{-1}(U)=\{ Y’\in\pi_m(Z): p_m((\pi_m|Z)^{-1}(Y’))\in U\}=$$ $$\{Y’\in\pi_m(Z): (\pi_m|Z)^{-1}(Y’)\in p_m^{-1}(U)\}=(\pi_m|Z)(p_m^{-1}(U))$$
is open in  $\pi_m(Z)$. Conversely, assume that the map $h$ is continuous. Let $V$ be any open subset of $Z$. Then a set
$$(\pi_m|Z)(V)=\{Y’\in \pi_m(Z): h(Y’)\in V\}=h^{-1}(V),$$ 
is open in $\pi_m(Z)$.
Finally we remark that this condition does not always hold because we can easily construct an embedding $g:\Bbb R\to\Bbb R^3$ such that projection $\pi_{3}|g(\Bbb R)$ is injective, but not open onto image. 
