I have the following:
\begin{align} \frac{\partial }{\partial t} f(x,t)&=g(f(x,t),t) \tag 1\\ f(x,0)&= x \tag 2 \end{align} where $g:\mathbb R^{n+1}\to \mathbb R^n$
The domain and range for $f$ is not stated, but I assume $f:\mathbb R^{n+1}\to \mathbb R^n$?
However, in the RHS we have a function composition of $g$ and $f$, i.e. $g\circ f$. But the composition is not valid if $f:\mathbb R^{n+1}\to \mathbb R^n$ according to the definition (Wikipedia):
The functions $f:X \rightarrow Y$ and $g:Y\rightarrow Z$ are composed to yield a function... The resulting compositie function is denoted $g\circ f : X \rightarrow Z$, defined by $(g\circ f)(x) = g(f(x))$.
How can I "see" the domain and range for $f$ and $g\circ f$ from $(1)-(2)$?
Since domain of $g$ is $\Bbb{R}^{n+1}$, this means the vector $(f(x,t), t)$ must be in $\Bbb{R}^{n+1}$. Consequently, $f(x,t) \in \Bbb{R}^n$. So the co-domain for $f$ is $\Bbb{R}^{n}$. From the second equation given, we also get that $x \in \Bbb{R}^n$.
But observe that the input vector $(x,t)$ will now be in $\Bbb{R}^{n+1}$.
Finally we can say that $f:\Bbb{R}^{n+1} \rightarrow \Bbb{R}^n$.
The composition map $g \circ f$ is NOT defined. Note that $g(f(x,t), t)$ is NOT the composition of $g$ and $f$ because there is an extra argument $t$ in the input for $g$.
Example Let $g:\Bbb{R}^3 \rightarrow \Bbb{R}^2$ be defined as $g(p,q,r)=(p+q,r)$ and let $f:\Bbb{R}^3 \rightarrow \Bbb{R}^2$ be defined as $f(a,b,c)=(a^2,b+c)$.
Let $\mathbf{x}=(x_1,x_2)$ \begin{align*} g(f(\mathbf{x},t), t)&=g(f(x_1,x_2,t)\,, \,t )\\ &=g((x_1^2,x_2+t), \, \, t)\\ &=g(x_1^2,\,\, x_2+t, \,\, t)\\ &=(x_1^2+x_2+t, \,\, t) \end{align*}