I cant understand this when I have a system thar compress the input as follows $$y(t)=x(2t)$$ Now to check for time invariance now I should input $x(t)$ to get $$y_1(t) = x(2t)$$ Then I should input $x(t-t_\circ)$ to get $$y_2(t) =x(2(t - t_\circ)) = x( 2t - 2t_\circ)$$ Now checking the output shifted we get $$y_1(t-t_\circ) = x(2(t - t_\circ)) = x( 2t - 2t_\circ)$$ And since $y_2(t) =y_1(t-t_\circ)$ then this is time invariance though I know from my book that this last step of shifting $y_1(t)$ is not correct but I cant understand why?!!
If $y(t)=x(2t)$ then for time-invariance the response to the input signal $x(t-t_0)$ should be $y(t-t_0)=x(2(t-t_0))$. Let's call the new input $z(t)=x(t-t_0)$. The corresponding output is $y(t)=z(2t)=x(2t-t_0)\neq x(2(t-t_0))$, which shows that the system is indeed time-varying.