Piecewise smooth curve has a smooth representative

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Let $\gamma : [0,1] \rightarrow \mathbb{R}^{2}$ be continuous path such that $\gamma |_{[0,1/2]}$ and $\gamma |_{[1/2,1]}$ are smooth. Does there exists a continuous bijection $\varphi :[0,1] \rightarrow [0,1]$ such that $\gamma \circ \varphi$ is smooth on the whole $[0,1]$ and $\varphi(0)=0,\varphi(1)=1$?

Another question that I have is: Can a smooth path $\gamma :[0,1] \rightarrow \mathbb{R}^{2}$ have infinitely many self-intersections?

Greetings, Phil

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Only a partial answer for the second question. Let be $$ \begin{gathered} \gamma :\left[ {0,1} \right] \to \mathbb{R}^2 \hfill \\ t \to \left( {x(t),y(t)} \right) \hfill \\ \end{gathered} $$ such that $$ x(t) = \left\{ \begin{gathered} 0\,\,\,\,if\,\,\,t = 0 \hfill \\ t^3 \cos \left( {\frac{1} {t}} \right)\,\,\,if\,\,\,0 < t \leqslant 1 \hfill \\ \end{gathered} \right. $$ $$ y\left( t \right) = \left\{ \begin{gathered} 0\,\,\,\,if\,\,\,t = 0 \hfill \\ 2t^3 + t^3 \sin \left( {\frac{1} {t}} \right)\,\,\,if\,\,\,0 < t \leqslant 1 \hfill \\ \end{gathered} \right. $$ Then the curve is smooth. Moreover it seems to have the required property as shown in the pic. However, so far,I I'm not able to prove it analytically. enter image description here

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Hint for first question: Find a smooth function $\varphi_1$ on $[0,1/2]$ such that $\varphi_1(0)=0, \varphi_1(1/2)=1/2,$ $\varphi_1$ is strictly increasing, and all derivatives of $\varphi_1$ at $1/2$ are $0.$ Something like $1/2-ce^{1/(t-1)}$ should do it. Hook it up with a similar $\varphi_2$ on $[1/2,1]$ in the appropriate way. Join these curves together to give $\varphi$ on $[0,1].$

For the second question, I'll build a curve in parts. Define

$$\gamma (t)= \begin{cases}(1,0)-e^{1/(t-1)}(1,\sin(1/(t-1))), \,\,t\in [0,1)\\ (1,0),\,\,t=1\\ (1-e^{1/(1-t)},0),\,\, t\in (1,2]\end{cases}$$

Note that all left and right derivatives of $\gamma$ at $1$ are $0.$ Thus $\gamma$ is smooth. Now $\gamma(0)$ is just some point in $\mathbb R^2$ we don't care too much about. But note that for $t_n=1-1/(n\pi),n=1,2,\dots,$ the $\gamma(t_n)$ are distinct points on the $x$-axis. Since the final leg of $\gamma$ just sends us to the left of $0$ along the $x$-axis, we have the desired infinite collecton of self-interesection points of $\gamma.$