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?

If $V$ and $W$ are finite-dimensional (real) vector spaces of the same dimension, then a linear transformation $L:V \to W$ is injective if and only if $L$ is surjective.
Consequently, if $f:M \to N$ is a smooth mapping of manifolds of the same dimension, the following are equivalent:


*

*$f$ is an immersion.

*$df_{x}:T_{x}M \to T_{f(x)}N$ is injective for every $x$ in $M$.

*$df_{x}:T_{x}M \to T_{f(x)}N$ is surjective for every $x$ in $M$.

*$f$ is a submersion.
As paf notes, asking about injectivity (or surjectivity) of $f:M \to N$ itself is another matter. For example:


*

*The real exponential map is an injective immersion, but not surjective.

*The squaring map on the unit circle is a surjective immersion, but not injective.

*The complex exponential map is an immersion, but neither injective nor surjective.
