If $f: A \to B$ is given by $f(x,y) = (x,y,0)$, where $A$ is an open disc in $\mathbb{R^2}$ and $B=\{{(x,y,0)}\in \mathbb{R^3} \mid (x,y) \in A\}$ is there a way to show that both $f$ and $f^{-1}(x,y,0) = (x,y)$ are continuous without using the formal epsilon-delta definition of continuity?
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You can use the fact that in general, if$f: A \rightarrow C$ and $g: B \rightarrow D$ are continuous then $f \times g: (A \times B) \rightarrow (C \times D)$ is continuous, and that projections are continuous. – Jair Taylor Mar 18 '20 at 21:05
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Let $U \subseteq \mathbb{R}^n$ and $g:U \to \mathbb{R}^m$ given by $g(\mathbf{x}) = (g_1(\mathbf{x}), g_2(\mathbf{x}), \dots, g_m(\mathbf{x}))$, where $g_1,g_2,\dots,g_m : U \to \mathbb{R}$ are its coordinate functions. Then, $g$ is continuous if and only if each $g_i$ is continuous. Of course, this result need to be proven with the $\varepsilon$-$\delta$ definition.
azif00
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The open sets of $B$ are precisely $\{(x,y,0):(x,y)\in U\}$ for open sets $U$ of $A$. In other words, $V$ is an open set of $B$ if and only if $f(U)=V$ for some open set $U$ of $A$.
Ben W
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