General Topology – Graph of a Continuous Function is Locally Euclidean

general-topologymanifolds

I am working through Lee's "Intro to Topological Manifolds," and in the third chapter the subspace topology is introduced. After showing that subspaces of Hausdorff and second-countable spaces are Hausdorff and second-countable, we realize that since $\mathbb{R}^n$ satisfies both of these conditions, a subspace of $\mathbb{R}^n$ will be a manifold if it is also locally Euclidean; hence, to know if a subspace of $\mathbb{R}^n$ is a manifold, we need only check that it is locally Euclidean. (A manifold is a Hausdorff, second-countable, locally Euclidean topological space.)

A first example is that of a continuous map $f:U\subset \mathbb{R}^n \rightarrow \mathbb{R}^k$, where $U$ open; the claim is that the graph $$ \Gamma\left(f\right) = \left\{ \left(x,f\left(x\right)\right) \, \middle| \, x \in U \right\} \subset \mathbb{R}^{n+k} $$ is an $n$-manifold. As mentioned above, it's surely Hausdorff and second-countable.

Naturally the homeomorphism between $\Gamma\left(f\right)$ and $U$ that we seek is going to be $$ \Phi \left(x \right) = \left(x,f\left(x\right)\right) $$ and it remains to show that this is, indeed, a homeomorphism. It is certainly a bijection. We can see that $\Phi ^{-1}\left(x,f\left(x\right)\right) = x $ so that $$ \Phi^{-1} = \pi_1|_{\Gamma\left(f\right)}$$ which is the restriction of a continuous map, and is thus continuous.

I am struggling to show that $\Phi$ itself is continuous. Note that the author has not yet introduced product topologies, and I am meant to work only with subspaces. The claim is that "It is continuous because $f$ is," and I am trying to prove this. My proof seems much messier than that concise statement suggests. Am I missing something simple here?

My proof

Let $V\subset \Gamma\left(f\right)$ be open, so $V = W\cap \Gamma\left(f\right)$ with $W\subset \mathbb{R}^{n+k}$ by definition of the subspace topology on $\Gamma\left(f\right)$. We want to show that $\Phi^{-1}\left(V\right)$ is open in $U$. Let $x \in \Phi^{-1}\left(V\right)$. Then $\Phi\left(x\right) \in V = W \cap \Gamma\left(f\right)$, so $\Phi\left(x\right) \in W$ and there exists some ball neighborhood of $\Phi(x) = \left(x,f\left(x\right)\right)$ contained in $W$, namely $B_1 = B_{\epsilon}\left(\left(x, f\left(x\right)\right)\right) \subset W$.

(I will now show that there is a neighborhood of $x$ contained in $\Phi^{-1}\left(V\right)$ that maps within $B_1$, proving the openness of the former and thus continuity of $\Phi$. This will utilize continuity of $f$ and properties of the Euclidean metric.) Since $f:\mathbb{R}^n \rightarrow \mathbb{R}^k$ is continuous, there is a $\delta' >0$ such that $$d\left(x', x\right) < \delta' \implies d\left(f\left(x'\right), f\left(x\right)\right) < \epsilon / \sqrt{2}, $$ where the $d$ indicate the metrics on $\mathbb{R}^n$ and $\mathbb{R}^k$ respectively. Then let $\delta = \min \left\{\delta', \epsilon/\sqrt{2}\right\}$. Then if $ d\left(x', x\right) < \delta$, we have

$$ d\left(\left(x', f\left(x'\right)\right), \left(x, f\left(x\right)\right) \right)^2 = d\left(x', x\right)^2 + d\left(f\left(x'\right), f\left(x\right)\right)^2 < \left(\frac{\epsilon}{\sqrt{2}}\right)^2 = \epsilon^2 $$ which says that $$ d\left(\Phi\left(x'\right), \Phi\left(x\right)\right) < \epsilon $$ and so we have shown that $$x' \in B_{\delta}\left(x\right) \implies \Phi\left(x'\right) \in B_1. $$ Since $\Phi\left(x'\right) \in \Gamma\left(f\right)$ obviously, we have further that $$x' \in B_{\delta}\left(x\right) \implies \Phi\left(x'\right) \in B_1 \cap \Gamma\left(f\right) \subset W\cap \Gamma\left(f\right) = V. $$ Hence, $B_{\delta}\left(x\right)$ is an open neighborhood of $x$ in $\Phi^{-1}\left(V\right)$ that maps inside of $V$, so $\Phi^{-1}\left(V\right)$ is open and $\Phi$ is continuous. $\blacksquare$

Best Answer

Lee's sometimes leaves the reader to fill in the blanks for routine checks. In example 3.20, which references example 3.14 - he does not argue at all that either $\Phi$ or $F$ are continuous. He states it as a fact, so the reader is left to assume that it should be straightforward to show. You'll notice that your argument, although longer than a few words, is pretty straightforward, so I wouldn't fret about the length.

I think your argument is overall fine, sans some nitpicks. Your argument boils down to:

  1. Let $x \in \Phi^{-1}(V)$.
  2. By (1), $\Phi(x) = (x,f(x)) \in V$.
  3. By (2) and since $V$ is open, $(x,f(x))$ has an open neighborhood $B_{\epsilon}((x,f(x)) \subset V$.
  4. Take any point $x' \in U$. Use the continuity of $f$ to find some $\delta$ that bounds the distances of $x$ and $x'$ such that the distance between $f(x)$ and $f(x')$ is bounded by $\frac{\epsilon}{\sqrt{2}}$.
  5. Realize the square distance between $\Phi(x)$ and $\Phi(x')$ as $d(x,x')^2+d(f(x),f(x'))^2$.
  6. Notice that if:
    1. $\delta < \frac{\epsilon}{\sqrt{2}}$, then $d(x,x')^2+d(f(x),f(x'))^2 < \epsilon^2$.
    2. $\delta > \frac{\epsilon}{\sqrt{2}}$, then picking a point $x''$ such that $d(x,x'') < \frac{\epsilon}{\sqrt{2}}$ gives $d(x,x'')^2, d(f(x),f(x''))^2 < \epsilon^2$
  7. By (6), we see that $$d(x,y)<\min\{\delta', \frac{\epsilon}{\sqrt{2}}\}\implies \Phi(y) \in B_{\epsilon}(\Phi(x)) \implies \Phi(y) \in V.$$ In other words, $y \in \Phi^{-1}(V)$.
  8. By (7), $B_{\min\{\delta', \frac{\epsilon}{\sqrt{2}}\}}(x) \subset \Phi^{-1}(V)$, so $\Phi^{-1}(V)$ is open. Thus, $\Phi$ is continuous.
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