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What shown below is a reference from the text Analysis on Manifolds by James Munkres. If you know the definition of manifold and of manifold boundary you can only read the last lemma in the image.

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So now let be $\alpha:[a,b]\rightarrow\Bbb R^n$ a $C^r$ function. Well since the interval $[a,b]$ is diffeomorphic to the interval $[0,1]$ without loss of generality we can suppose that $a=0$ and $b=1$. So in this case the map $\alpha|_{[0,1)}$ is a local patch of $M:=\alpha\Big[[0,1]\Big]$ defined in a set that is open in $H^1$ but not in $\Bbb R$ so that the point $\alpha(0)$ is a boudary point. Now let be $\phi:[0,1)\rightarrow(0,1]$ the diffeomorphism given by the equation $$ \phi(x):=1-x $$ for any $x\in[0,1)$ so that the map $\alpha\circ\phi:[0,1)\rightarrow M$ is a local patch of $M$ defined in a set that is open in $H^1$ but not in $\Bbb R$ so that the point $\big(\alpha\circ\phi\big)(0)=\alpha\big(\phi(0)\big)=\alpha(1)$ is a boundary point. However if we put $$ \phi(x):=\frac 1 2-x $$ then the point $\alpha\Big(\frac 1 2\Big)=\alpha(\frac 1 2-0)=\alpha\big(\phi(0)\big)=\big(\alpha\circ\phi\big)(0)$ was a boundary point too and this very strangerly because the map $\alpha|_{(0,1)}$ is a local patch about $\alpha\Big(\frac 1 2\Big)$ defined in an open set of $\Bbb R$ and the point $c$ of the lemma $24.2$ implies that the point $a$ can not be true as you can see in the proof and in any case the boundary and the interor of a manifold are disjoint directely by its definition. So I am sure there is something wrong in my arguments but I do not see it and so I ask to find it. So is correct the proof I gave to show that $\alpha(1)$ is a boundary point? So could someone help me, please?

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  • $\begingroup$ You're not paying attention to domains. How is $\alpha(1/2)$ a boundary point? $\endgroup$ – Ted Shifrin Feb 23 at 17:57
  • $\begingroup$ @TedShifrin The problem is that $\alpha(1/2)=\alpha(1/2-0)=\alpha(\phi(0))$ and it seems to me that $\alpha(\phi(0))$ is a boundary point because the map $\alpha\circ\phi$ is defined in a right neigborhood of zero. So what can you say about? Where am I getting wrong? $\endgroup$ – Antonio Maria Di Mauro Feb 23 at 18:02
  • $\begingroup$ By your logic, any point of $\Bbb R$ is a boundary point if you take a chart with the right domain. Just take $\alpha\colon [a,b)\to\Bbb R$, $\alpha(x)=x$, and you're going to tell me that $a$ is a boundary point of $\Bbb R$. Read his definition very carefully. $\endgroup$ – Ted Shifrin Feb 23 at 18:09
  • $\begingroup$ @TedShifrin Okay, and clerly this is impossible. But why my arguments are incorrect with respect Munkres formalism? So how prove that $\alpha(b)$ is a boundary point? I am wrong when I use the diffeomorphism $\phi$ to create a local patch about $\alpha(1)$? Could you explain this, please? Forgive my confusion. $\endgroup$ – Antonio Maria Di Mauro Feb 23 at 18:13
  • $\begingroup$ Munkres is very careful to define a boundary point to be a point that is not an interior point — not a point that is the image of a boundary point under some chart. (Definitions can be done in different ways, but Munkres is very careful here.) $\endgroup$ – Ted Shifrin Feb 23 at 18:15

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