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I read the following comment in a book I'm reading:

Let $J\subset \mathbb R$ be an open interval. A differentiable path $f:J\to \mathbb R^n$ is an immersion if and only if its velocity vector is non-zero for every $t\in J$.

The general definition of immersion says if $U\in \mathbb R^n$ is open and $f:U\to \mathbb R^n$ is a differential function then for every $x\in U$ we have $f'(x)$ is injective. I don't know how to use this definition to prove this equivalence.

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    The velocity vector at $t=t_0$ is $f'(t_0)$ — or $f'(t_0)(1)$, if you insist on inputting a vector in $\Bbb R$. – 2017-01-11
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    A linear transformation with one-dimensional domain is injective if and only if it is non-zero. (Hi Ted. :) – 2017-01-11
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    Ted's parenthetical remark reminds me of Dieudonne's criticism of how most 1D calculus classes teach that the derivative at a point is a number rather than a linear transformation. His complaint starts with something like: "This slavish subservience to the shibboleth of numerical interpretation at any cost..." (Hi, Ted. :) – 2017-01-11

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The velocity vector for a curve $f$ at position $t$ is $$ f'(t) = df(t)[\frac{\partial}{\partial t}] $$ If the left is nonzero, the right must be as well, and hence the multiplier on the right, $df(t)$ must be nonzero. Since a linear map from $\Bbb R^1$ to $\Bbb R^2$ is injective if and only if it's nonzero, we're done.

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    Of course we never know what level of notation the OP has learned ... But it's fun old time reunion here, John, @AndrewD.Hwang :) – 2017-01-11