Why a map is equivariant if and only if it is equivariant in the infinitesimal version?

Solution 1:

If $G$ is not connected then they are certainly not equivalent, since the infinitesimal statement only depends on the action of the connected component of the identity in $G$. For instance, if $G$ is $0$-dimensional, the infinitesimal statement is always trivially true.

If $G$ is connected, though, the infinitesimal statement does imply that $f$ is equivariant. To prove this, note that for $X\in\mathfrak{g}$, $t\mapsto \exp(tX)^i\cdot m$ is the flow along the vector field $\underline{X}^i$ starting from $m$. If $f$ is infinitesimally equivariant, then it preserves these flows, and so it preserves the action of elements of $G$ of the form $\exp(X)$. When $G$ is connected, it is generated by the image of the exponential map, and so it follows that $f$ is actually equivariant.