I am trying to prove the following statement:
If the functor $L_E$ exists,
$(iii)$ for any map $g: X \to Y$ where $Y$ is $E_*$-local, there is a unique map $\tilde{g}: L_E X \to Y$ such that $\tilde{g} \eta_X = g.$
Here is what is given to me in the paper "Localization with Respect to Certain Periodic Homology Theories."
(Note there were a typo in $(iii)$ but I corrected it in my statement)
Still, I do not know how to prove this statement, could anyone help me please?
Let $F$ be the fiber of the map $\eta_X: X \to L_E X$, so that we have an exact sequence $$\cdots \rightarrow [\Sigma F, Y] \rightarrow [L_E X, Y] \xrightarrow{(\eta_X)^*} [X, Y] \rightarrow [F, Y] \rightarrow \cdots.$$ We want to show that the map $(\eta_X)^*$ is a bijection. For this it suffices to show that $[F, Y] = [\Sigma F, Y] = 0$.
Since $\eta_X$ is an $E$-equivalence, $F$ and thus also $\Sigma F$ are $E_*$-acyclic using the long exact sequence in $E_*$-homology. Then, since $Y$ is $E_*$-local, $[F, Y] = [\Sigma F, Y] = 0$ as desired.
We conclude that $(\eta_X)^*: [L_E X, Y] \to [X, Y]$ is a bijection; i.e., given any $g \in [X,Y]$, there exists a unique $\tilde{g} \in [L_E X, Y]$ such that $\tilde{g} \circ \eta_X = g$.