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.
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?