Let $f, g:D^2\longrightarrow \mathbb R$ be continuous such that $f|_{\partial D^2}=g|_{\partial D^2}$. I want to show there exists $F:S^2\longrightarrow \mathbb R$ continuous such that $F|_{H_{+}}=f$ and $F|_{H_{-}}=g$.
Sketch of a Proof: Define $F:S^2\longrightarrow \mathbb R$ setting: $$F(x)=\left\{\begin{array}{ccc}f(x)&\textrm{if}&x\in H_{+}\\ g(x)&\textrm{if}&x\in H_{-}\end{array}\right..$$ If $x\in H_+\cap H_-=S^1=\partial D^2$ then $f(x)=g(x)$ for $f|_{\partial D^2}=g|_{\partial D^2}$ and therefore, by the gluing lemma, $F$ is continuous since $H_+$, $H_-$ are closed and $S^2=H_+\cup H_-$. From the definition of $F$ we see $F|_{H_{+}}=f$ and $F|_{H_{-}}=g$.
Problem: Rigorously, the domain of $f$ and $g$ are $D^2$ not $H_{\pm}$ so how to overcome this? I know $H_+$ and $H_-$ are homeomorphic to $D^2$ but I don't know how to argue from this. Can anyone help me?
Notation: $H_+$ and $H_-$ are the upper and lower hemispheres of the unit sphere $S^2$. $D^2$ is the closed unit disc.
The obvious projection $\pi: \mathbb R^3 \to \mathbb R^2 \cong \mathbb R^2 \times \{0\}$ is a homeomorphism on its image when restricted to $H_\pm$.
The argument uses that the graph of a continuous function $\varphi: X \to Y$, i.e. the set $\{(x,\varphi(x)) \mid x \in X\} \subset X \times Y$, is homeomorphic to the domain via the projection $(x,y) \mapsto x$ with inverse $x \mapsto (x,\varphi(x))$. This is of course the bit that requires proof!
And $H_\pm$ are graphs of the continuous functions $s_\pm(x,y) = \pm\sqrt{1-x^2-y^2}$. This too should be verified.
All that's left is to define $\tilde f = f \circ \pi$, do the same thing for $g$, and to say something about being done.