$(a)$ Let $\sigma :\Bbb R \to \Bbb R$ be $\sigma(t) = t^2$. Is $(\sigma , \Bbb R)$ a coordinate chart on $\Bbb R$?
$(b)$ Let $\phi :\Bbb R \to \Bbb R$ be $\phi(t) = t^3$. Does this gives a smooth manifold structure on $\Bbb R$?
$(c)$ Determine if the function $f: \Bbb R \to \Bbb R^2$ given by $f(t) = (t^2,t^3)$ is an immersion at $0$.
$(a)$ $(\sigma , \Bbb R)$ is not a coordinate chart on $\Bbb R$ as it does not cover the -ve points.
$(b)$ It does not give a smooth manifold structure on $\Bbb R$, as $t \to t^3$ is a homeomorphism but it is not a diffeomorphism, as its inverse $x\mapsto x^{\frac 13}$ is not smooth.
$(c)$ $f$ is an immersion if $\forall p\in\mathbb{R}$, $df_p$ is injective. I have computed $Df_p=(2t,3t^2)$ . I have an intuition $df_p$ is injective at $0$.
Please verify the solution, and need help for part $(c)$.
For (a), you can argue that $\sigma$ is not injective, also. Your reasoning is correct, though.
In (b), for the atlas ${\cal A} = \{(\Bbb R, \phi)\}$, the only coordinate change is the identity map, which is differentiable, so ${\cal A}$ is contained in a maximal atlas, and the answer is yes. If you have not yet fixed differentiable strucutres in $\Bbb R$, it does not makes sense to ask if $\phi$ is differentiable. When you write $\phi:\Bbb R\to \Bbb R$, the first $\Bbb R$ is the topological manifold you're trying to endow with a differentiable structure, and the second $\Bbb R$ is the "model" of the chart.
For (c), since $f'(0)=(0,0)$, the derivative $df_{0}:T_0\Bbb R\to T_{(0,0)}\Bbb R^2$ is the zero map, hence $f$ is not an immersion.