Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin–Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{n}f^n(T)$. Then $\lambda_F(T) = T + \text{higherdegree terms}$, so the logarithm is invertible under function composition. We define $\operatorname{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\operatorname{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

5$\begingroup$ It's 1 to 1 in a neighborhood of 0, and so the identity is also. But as you say, it can't be true over all padic algebraic integers, even when restricted to positive valuation. Try to compute the radius of convergence of the log function. Note that all of this has an analogue over the complex numbers $\endgroup$– Dror SpeiserOct 23, 2021 at 6:57

3$\begingroup$ To echo something implicit in Dror's comment: the same issue arises when F is the multiplicative group law, so it might be best to understand that one first; in this case we have an explicit formula for the coefficients of log and exp, so the radius of convergence is easily computable. $\endgroup$– David LoefflerOct 23, 2021 at 8:17
1 Answer
The radius of convergence of any formal group $F$ (onedimensional, finite height) is $1$, in other words $L_F$ will converge at all $z\in\Bbb C_p$ with $v(z)>0$.
In particular, the logarithm is convergent at all torsion points of the formal group. What goes wrong, goes wrong with the exponential, whose series is not convergent far enough from the origin to reach any of the torsion points.

4$\begingroup$ As Lubin has commented somewhere else on this site, it is a remarkable property that the wellbehaved exponential (everywhere convergent, and homomorphism) and badlybehaved logarithm (finite radius of convergence, not a homomorphism) in complex analysis switch roles in $p$adic analysis, where the logarithm is better behaved than the exponential in the sense of having a larger radius of convergence (although both are $p$adically nice in being homomorphisms between their discs of convergence). $\endgroup$– KConradOct 30, 2021 at 22:48