Do we have $\mathbb{R} \times \mathbb{R}^{3}= \mathbb{R}^{4}$ or just $\mathbb{R} \times \mathbb{R}^{3} \simeq \mathbb{R}^{4}$ I came across this map in an exercise,
$$f\colon\left\{\begin{array}{rcl} \color{red}{\mathbb{R} \times \mathbb{R}^{3}} & \longrightarrow & \mathbb{R}^{3} \\ \left(t, \begin{pmatrix} x \\ y \\ z \end{pmatrix}\right) & \longmapsto & \begin{pmatrix}
x+y+z+t \\ x^{2}+y^{2}+z^{2}+t-2 \\ x^{3}+y^{3}+z^{3}+t^{2}
\end{pmatrix}. \end{array}\right.$$
The question is, do we have $\mathbb{R} \times \mathbb{R}^{3}= \mathbb{R}^{4}$ and thus the author wrote $\mathbb{R} \times \mathbb{R}^{3}$ with an ordered pair $\left(t, \begin{pmatrix} x \\ y \\ z \end{pmatrix}\right)$ for conveniance/the sake of clarity in a certain purpose. Or is $\mathbb{R} \times \mathbb{R}^{3}$ really different from $\mathbb{R}^{4}$ by definition, conceptually or structurally ? And thus this notation would be mandatory and we would only have $\mathbb{R} \times \mathbb{R}^{3} \simeq \mathbb{R}^{4}$.
I guess we have the equality for two reasons:

*

*Because we have $\mathbb{R}^{n}= \underbrace{\mathbb{R} \times \ldots \times \mathbb{R}}_{n \text{ times}}$ by definition, and this definition/notation suggests that we can do $$\color{red}{\mathbb{R}} \times \mathbb{R}^{3}= \color{red}{\mathbb{R}} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R}= \mathbb{R}^{4}$$


*As I remember we do have by definition $(x,y,z,t):= (x,(y,z,t))$ with a recursive definition, until we get to an ordered pair defined by $(a,b)= \{ a, \{ a,b \}\}$. Thus, since we have $\left(t, \begin{pmatrix} x \\ y \\ z \end{pmatrix}\right)=(t,x,y,z)$ which is an element of $\mathbb{R}^{4}$, thus we would have $\color{red}{(?)}$ in $$\mathbb{R} \times \mathbb{R}^{3}:= \mathbb{R} \times \{ (x,y,z) : x,y,z \in \mathbb{R} \}= \{ (t,(x,y,z)) : t,x,y,z \in \mathbb{R} \} \underset{\color{red}{(?)}}{=} \{ (t,x,y,z) : t,x,y,z \in \mathbb{R} \} := \mathbb{R}^{4}$$
 A: Hier are some aspects which might be helpful. This answer is based upon naive set theory, neither using axiomatic set theory nor formal logic. Nevertheless I think this approach is sufficient to gain some insight.
Sets usually carry additional structure
We have to be carefully, when considering
\begin{align*}
\mathbb{R} \times \mathbb{R}^{3}= \mathbb{R}^{4}\qquad\mathrm{vs.}\qquad \mathbb{R} \times \mathbb{R}^{3} \simeq \mathbb{R}^{4}
\end{align*}
Since for instance in OPs example
\begin{align*}
&f\colon\mathbb{R} \times \mathbb{R}^{3} \longrightarrow \mathbb{R}^{3} \tag{1}\\
&f(t,(x,y,z)) =\begin{pmatrix}
x+y+z+t \\ x^{2}+y^{2}+z^{2}+t-2 \\ x^{3}+y^{3}+z^{3}+t^{2}
\end{pmatrix}
\end{align*}
the sets $\mathbb{R}$ and $\mathbb{R}^3$ are not only sets, they are endowed with a vector space structure over the field $\mathbb{R}$, resp. with the field structure of $\mathbb{R}$ in order to do arithmetical operations.

This rather indicates the appropriate question in (1) is not about set-isomorphisms $\mathbb{R} \times \mathbb{R}^{3} \simeq \mathbb{R}^{4}$ and set-equality $\mathbb{R} \times \mathbb{R}^{3}=\mathbb{R}^{4}$ but vectorspace-isomorphisms between $\mathbb{R} \times \mathbb{R}^{3} \simeq \mathbb{R}^{4}$ and vectorspace-equality instead. We know from linear algebra that as vectorspaces they are isomorphic. Usually we don't consider them as being equal.

In the following we restrict considerations solely  to sets without additional structure and to set-isomorphisms.
The essential property of $n$-tupel is order
We can read in section 6, Naive Set Theory by P. R. Halmos:

*

*... This set is called the Cartesian product of $A$ and $B$; it is characterized by the fact that
\begin{align*}
A\times B=\{x:x=(a,b)\ \text{ for some }a\text{ in }A\text{ and for some }b\text{ in }B\}\tag{2}
\end{align*}
The Cartesian product of two sets is a set of ordered pairs (that is, a set each of whose elements is an odered pair), and the same is true of every subset of a Cartesian product. It is of technical importance to know that we can go in the converse direction also: every set of ordered pairs is a subset of the Cartesian product of two sets.

We observe the essential properties of elements here are ordered pairs. So, we have some kind of ordering and some kind of two-ness. This indicates we can talk about set-isomorphims between $\mathbb{R} \times \mathbb{R}^{3}$ and $\mathbb{R}^{4}$, but not about equality, since
\begin{align*}
\color{blue}{\mathbb{R}^{4}=\{(a,b,c,d): a,b,c,d\in\mathbb{R}\}}
\end{align*}
is characterized via $4$-tupel, while in
\begin{align*}
\color{blue}{\mathbb{R}\times \mathbb{R}^{3}=\{(a,(b,c,d)): a\in\mathbb{R},(b,c,d)\in\mathbb{R}^3\}}
\end{align*}
we have ordered pairs.
Definition of $n$-tupel
There is more than one way to define $n$-tupel. One is given by OP when ordered pairs $(a,b)$ are defined as set
\begin{align*}
(a,b):=\{\{a\},\{a,b\}\}
\end{align*}
We can in the same way define $4$-tupel as
\begin{align*}
(a,b,c,d):=\{\{a\},\{a,b\},\{a,b,c\},\{a,b,c,d\}\}
\end{align*}
and note, the essential property of $n$-tupel is order, not necessarily the way it is defined. Each useful definition has just to incorporate this essential property.
Using for instance this definition we again see, that as sets the $4$-tupel $(a,b,c,d)$ and the ordered pair $(a,(b,c,d))$
\begin{align*}
(a,b,c,d)&:=\{\{a\},\{a,b\},\{a,b,c\},\{a,b,c,d\}\}\\
(a,(b,c,d))&:=\{\{a\},\{\{a\}, (b,c,d)\}\}=\{\{a\},\{\{a\},\{\{b\},\{b,c\},\{b,c,d\}\}\}\}
\end{align*}
are different. But there is a set-isomorphism $\mathbb{R}\times \mathbb{R}^{3}\simeq \mathbb{R}^{4}$
Conclusion:
We have a set-isomorphism $\mathbb{R}\times \mathbb{R}^{3}\simeq \mathbb{R}^{4}$ which is usually not considered to be an equality. Nevertheless sometimes it is convenient, as we have seen in OPs example of inductive definition of $\mathbb{R}^n=\mathbb{R}\times\mathbb{R}^{n-1}$ to identify these sets.
Note:

*

*Regarding OPs question, if the author when defining the function $f$ has done this for the sake of clarity and/or with a specific purpose. The answer yes seems plausbile, since we have a slightly different notation of variables, $t$ (time) on the one hand and $x,y,z$ (space) on the other hand.


*Another technical aspect is: We sometimes use the $j$-th projection map to select the $j$-th component of an $n$-tupel. In case of $\mathbb{R}\times \mathbb{R}^{3}$ we have
\begin{align*}
\pi_2(t,(x,y,z))=(x,y,z)
\end{align*}
whereas  in $ \mathbb{R}^{4}$ we have
\begin{align*}
\pi_2(t,x,y,z)=x
\end{align*}
A: To paraphrase Matthew

Sufficient unto the day is the rigor thereof.

In the context of the exercise that's the source of the question, whether those vector spaces are equal or isomorphic does not matter. The author's meaning is clear to the reader.
Requiring more formal precision  would make it harder to understand what the exercise is designed to teach.
As other answers and the long comment thread with @Smiley1000 show, there are other contexts where the distinction may be important.
