In the lecture notes by J.P. May, The geometry of iterated loop spaces, Chapter 5, formula (1), (2) and (10), a map $$ \phi: Hom_T(X,\Omega Y)\to Hom_T(SX,Y) $$ is defined. And a map $$ \eta_n=\phi^{-n}(1_{S^nX}):X\to \Omega^n S^nX. $$
Question: I do not quite understand what does $\eta_n$ mean? What means $\phi^{-n}(1_{S^nX})$?
$\phi$ is the map that appears in the standard adjunction between loop spaces and suspension: the suspension functor $S$ is the left adjoint of the loop space functor $\Omega$. This is a very slight abuse of notation, as the map $\phi$ depends on the spacse $X$ and $Y$, so it could actually be called $$\phi_{X,Y} : \hom_T(X, \Omega Y) \to \hom_T(SX, Y)$$ Being part of an adjunction, $\phi$ is a bijection, so it makes sense to talk about its inverse.
The map $1_{S^n X}$ is the identity of the iterated $n$th suspension of the space $X$: $S^n X = S(S(\dots S(X) \dots))$. We have $1_{S^n X} \in \hom_T(S^n X, S^n X)$; we can apply the inverse of the adjunction map to get $\phi_{S^{n-1}X, S^n X}^{-1}(1_{S^n}) \in \hom_T(S^{n-1} X, \Omega S^n X)$. You can continue to apply the adjunction to eventually get a map in $\hom_T(X, \Omega^n S^n X)$; this is the map that May denotes $\phi^{-n}(1_{S^n})$. Again, it's a slight abuse of notation, as the map $\phi$ is not the same at each step (first it's $\phi_{S^{n-1}X, S^n X}$, then $\phi_{S^{n-2}X, \Omega S^n X}$, etc.) but it's not really a big deal.
You can also view it this way: since $S$ is left adjoint to $\Omega$, $S^n$ is left adjoint to $\Omega^n$ (essentially what I just did), and $\eta_n$ is the unit of this adjunction at $X$.