exp(X)=A has a solution if A is sufficiently close to identity matrix Can anyone give hint for proving this? I think some kind of inverse thm argument would work. But I wasn't able to make it accurate... such as continuity of mapping $X \mapsto \exp(X)$ or this mapping has full rank near $A$ if $A$ is sufficiently close to identity...

Since $\exp(A) = I + A + O(A^2)$, the derivative of $\exp$ at $0$ is the identity map. The inverse function theorem then shows that $\exp$ is invertible in a neighbourhood of $I$.