confusion with differential of map I was told to prove $f(x)=x^3$ from $\mathbb R$ to $\mathbb R$ is not an immersion nor a submersion, and now I'm really confused. As a function $f^\prime$ is not injective or surjective, but I'm supposed to check the differential:
Let $f:\mathbb R\to\mathbb R$ be given by $x\mapsto x^3$. I read $df_x:T_x\to T_{f(x)}$ is given by $\gamma ^\prime (0)\mapsto (f\circ \gamma)^\prime (0)$. But $(f\circ \gamma)^\prime(0)=f^\prime(\gamma(0))\gamma^\prime(0)$ no? And since the domain is $T_x$ then $\gamma(0)=x$ so it looks like $df_x$ is just multiplication by $f^\prime (x)$.
This means $df_x$ is just multiplication by $f^\prime (x)$. Okay, so a smooth function $\mathbb R\to\mathbb R$ is an injection iff it's an immersion iff it's derivative is nonzero? That doesn't make sense geometrically to me because I was taught they are "dual" to each other. Am I doing something wrong?

You're right. For functions $f:\Bbb R\to \Bbb R$, the differential of $f$ at a point $a$ is the multiplication by $f'(a)$, since, when $h\to 0$,
$$f(a+h) = f(a)+f'(a)h+o(h).$$
And your explanation with tangent vectors is also correct.
Thus, $f$ is an immersion at $a$ iff $f'(a)\neq 0$ (and this is also equivalent to $f$ being a submersion).But this has nothing to do with injectivity of $f$, which is a global notion, contrary to being an immersion at a single point. However, if $f$ is an immersion at every point, then $f$ is strictly incresing or strictly decreasing, hence injective.
