rt.representation theory - How to show the matrix exponential is onto? And, how to create a powerseries for log that works outside B(I,1)

My recollection is that Rossmann's book on Lie groups has a detailed discussion of the exponential map and surjectivity issue. Matrix exponential map is equivariant under conjugation,

$$exp(gXg^{-1})=gexp(X)g^{-1},$$

and, as Robin has already remarked, one can easily check that a matrix in Jordan normal form is in the image of $exp: M_n(mathbb{C})to GL_n(mathbb{C}),$ establishing surjectivity for $G=GL_n(mathbb{C}).$

As for your second question, you can always (a) rescale (b) shift by scalar matrices and re-center the $log$ series:

$$(a) exp(nB)=exp(B)^nquad (b) exp(B+lambda I_n)=e^{lambda}exp(B).$$

For topological reasons, there isn't a canonical formula for $log$ that works locally everywhere.

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Warning:. Exponential map is not always surjective. The following family of matrices is not in the image of the exponential map from the Lie algebra $mathfrak{g}=mathfrak{sl}_2(mathbb{C})$ (traceless $2times 2$ matrices) to the Lie group $G=SL_2(mathbb{C})$ (unit determinant $2times 2$ matrices):

$$h_a=begin{bmatrix}
-1 & a\
0 & -1
end{bmatrix}, ane 0.$$

No preimage in $M_2(mathbb{C})$ of $h_a$ under $exp$ can have trace $0$. Indeed, if $exp(X_a)=h_a$ then the eigenvalues of $X_a$ must be $pi i+2pi n, -pi i - 2pi m (n, min mathbb{Z})$ and the trace condition implies that $n=m,$ so $X_a$ has distinct eigenvalues, hence it is diagonalizable, but $h_a$ is not — contradiction.