1 parameter subgroups and Lie groups I was just reading some lecture notes (that are not online available unfortunately) on Lie groups and found that sometimes the author just says if he wants to prove something for all Lie group elements that it is sufficient to consider $1$ parameter subgroups $e^{t \eta}$ where $\eta \in \mathfrak{g}$ is an arbitrary element in the corresponding Lie algebra.
More precisely, he wants to show that $f(g) = h(g)\tag2$ for smooth functions $f,h:G \rightarrow \mathbb{R}$
then he says that this is the case if and only if
$$\frac{d}{dt}\bigg|_{t=0} f(e^{t \eta}) = \frac{d}{dt}\bigg|_{t=0} h(e^{t\eta}). \tag1$$
Under which conditions is equation $(2)$ equivalent to equation $(1)$?
 A: You need the functions $f$ and $h$ to respect the group structure in some way.
If $f\equiv0$ and $h$ is a nontrivial smooth function whose support doesn't contain the identity, then your condition is true but $f=h$ is false.
First of all, you need to assume that the exponential of the Lie algebra is dense in the Lie group.
Otherwise considering exponentials only will not suffice for drawing conclusions everywhere, and one can use a construction similar to what I gave above.
The exponential of the Lie algebra can only be dense if the group is connected, and for compact groups this condition is also sufficient.
A natural way to guarantee equivalence is to assume $f$ and $h$ to be homomorphisms, at least on each one parameter subgroup.
This is sufficient, not necessary; see what happens if you let $f$ and $h$ be multiples of a fixed function with nonzero differential at the identity.
If $f$ is a smooth homomorphism from the one parameter subgroup spanned by $\eta\in\mathfrak g$ to the additive group $\mathbb R$, then $t\mapsto f(e^{t\eta})$ is linear, so it is uniquely determined by its derivative at any point.
If $f$ is a smooth homomorphism to the multiplicative group $\mathbb R\setminus\{0\}$, then $f(e^{(t+s)\eta})-f(e^{t\eta})=f(e^{t\eta})[f(e^{s\eta})-1]$, so
$$
\left.\frac{d}{dt}f(e^{t\eta})\right|_{t=t_0}
=
f(e^{t_0\eta})
\left.\frac{d}{dt}f(e^{t\eta})\right|_{t=0}.
$$
Again, such a function is uniquely determined by its derivative at a point.
Check your lecture notes.
There must be some assumptions about the group and the functions under consideration.
