$\tanh (x+y)=\tanh (z)\rightarrow x+y=z?$

Question

Why is this true? I just read this in a relativity theory textbook without any proof. Is that really that obvious?

Answer 1

$$ \frac{\mathrm{d}}{\mathrm{d}x}\tanh(x)=\operatorname{sech}^2(x)\gt0 $$ So $\tanh(x)$ is strictly increasing, hence injective.

Answer 2

Using the definition and a simple calculation shows

$\tanh(a)=\tanh(b)$ iff $e^{a-b}=e^{b-a}$ iff $a=b$