Using the open sets definition of continuity to show that $f : S^2 → R^2$ is continuous. I have to show that a specific given function $f : S^2 → R^2$ such that $(x, y, z) \mapsto (a, b)$, where $a$ and $b$ are two compositions of continuous functions in $x$, $y$ and $z$ that are also given, is continuous.  I know I have to show that the preimage in $S^2$ of any open subsets of $R^2$ is open. My $a$ is such that $(x, y, z) \neq$ $(0, 0, 0)$, or it would be undefined, and my $b$ is such that $(x, y, z) \neq(1, 1, 1)$, or it would be undefined, but since $S^2 =\{(x, y, z)  |  x^2+y^2+z^2 = 1\}$, I can see that this would not be an issue.  Other than referencing the fact that the composition of continuous functions is continuous, is there a standard way to approach a question like this?  I have seen examples where we map from $R^3$ to $R$, but nothing like this.  Any help would be appreciated!
 A: Continuity is closed under composition, so $a, b$ are continuous. Cartesian product of continuous functions is continuous if the space is equipped with a conserving metric, and the Euclidean distance is in fact conserving. So this function is continuous.
A: Since $a$ and $b$ are defined using coordinates, we can define $F=(a,b)$ as an application from a subset $U$ of $\mathbb R^3$ into $\mathbb R^2$ and the question is to show that its restriction $f=F_{|S}$ from $S=S^2$ into $\mathbb R^2$ is continuous.
Since $a$ and $b$ are composition of continuous functions, they are continuous (as applications from $U$ into $\mathbb R$). Therefore, since $\mathbb R^2$ has the product topology, $F=(a,b)$ is continuous.
Now, since the topology of $S$ is the topology induced on $S$ by the topology of $\mathbb R^3$, the restriction $f=F_{|S}$ is continuous.
A: In my mind the easiest way to show that a function is continuous is by showing that it is given by composing continuous functions or that it is obtained from universal properties.
In this specific instance note that the universal property of the product states that to give a continuous map into $\Bbb R^n$ amounts to giving $n$ (possibly distinct) maps into $\Bbb R$. Hence it suffices to show that the maps $a: \Bbb S^2 \rightarrow \Bbb R$ and $b: \Bbb S^2 \rightarrow \Bbb R$ are continuous. Let me discuss the first one, the second is completely analogous.
Let $i: \Bbb S^2 \subseteq \Bbb R^3$ denote the inclusion of the sphere into the ambient space. The function $v:\Bbb R^3 \rightarrow \Bbb R^3, (x,y,z) \mapsto (\vert x \vert,\vert y\vert, \vert z \vert)$ is continuous by the universal property of $\Bbb R^3$ and the fact that $\vert-\vert: \Bbb R^3 \rightarrow \Bbb R$ is continuous. The function $p:\Bbb R^3 \rightarrow \Bbb R, (x,y,z) \mapsto x+y+z$ is continous. We obtain $a$ as the composite $p\circ v \circ i$, hence it is continous.
