I had defined two curves on $[0,1], R>1$ as $$\Gamma_1=\gamma_1+\gamma_2 \ \text{where}\\ \gamma_1(t)=Re^{2\pi it} \ \ \ \ \ \forall\ x\in[0,\frac{1}{2}]\\ \gamma_2(t)=4Rt-3R\ \ \ \ \ \forall\ x\in \big[\frac{1}{2},1\big] \\ and\ \Gamma_2(t)=i+\frac{1}{2}e^{2\pi it}\ \ \ \ \ \forall\ x\in[0,1]$$ then if i draw a figure then these two curves are homotopic but how can i show it explicitly. One idea in my mind is to use straight line homotopy but i am not sure about it.
Any type of help will be appreciated. Thanks in advance.
One idea is to take straight line homotopy as follows : $$\phi:[0,1]*[0,1]\rightarrow \mathbb R$$ as $\phi(s,t)=(1-s)\Gamma_1(t)+s\Gamma_2(t)$ and in order to show it is continuous at any $(s_0,t_0)\in [0,1]*[0,1]$ fix an $\epsilon>0$ be any and $m_1=\text{max}(|\Gamma_1(t_0)|,|\Gamma_2(t_0)|)$ and $s\in[0,1]$ s.t. $|s-s_o|<\frac{\epsilon}{4(m_1+1)}$ and let $m_2=\max(sup(|1-s|),sup|s|)$ where $"s"$ satisfies the above relation and $t\in U$ such that $\forall t\in U,\ |\Gamma_1(t)-\Gamma_1(t_0)|<\frac{\epsilon}{4(m_2+1)}$ and $|\Gamma_2(t)-\Gamma_2(t_0)|<\frac{\epsilon}{4(m_2+1)}$ (existence of such an U can be get by using continuity of $\Gamma_1$ and $\Gamma_2$). Hence$$|\phi(s,t)-\phi(s_0,t_0)|=|(1-s)\Gamma_1(t)+s\Gamma_2(t)-(1-s_0)\Gamma_1(t_0)+s_0\Gamma_2(t_0)|\\=|(1-s)\Gamma_1(t)+s\Gamma_2(t)-(1-s_0)\Gamma_1(t_0)-s_0\Gamma_2(t_0)+(1-s)\Gamma_1(t_0)-(1-s)\Gamma_1(t_0)+s\Gamma_2(t_0)-s\Gamma_2(t_0)|\\ \le|1-s||\Gamma_1(t)-\Gamma_1(t_0)|+|s-s_0||\Gamma_1(t_0)|+|s||\Gamma_2(t)-\Gamma_2(t_0)|+|s-s_0||\Gamma_2(t_0)|<\epsilon$$