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Why is cellularization the fiber of nullification for slice cells?

I'm a bit confused about the nullification functors that come up when constructing the slice tower in [HHR][1].

Let $\mathcal{A}$ be a set of compact objects in the $G$-equivariant stable homotopy category and let $\mathcal{A}_+$ denote the smallest subcategory that contains $\mathcal{A}$ and is closed under sums and taking cofibers. Then for a spectrum $X$ one can construct the $\mathcal{A}$-nullification $X\rightarrow P^{\mathcal{A}}X$. This has the universal property that it is $\mathcal{A}$-null, i.e. all maps from suspensions of objects in $\mathcal{A}$ into it are zero, and for all other $\mathcal{A}$-null spectra $Z$ and $k\geq 0$ the induced map $[\Sigma^k P^{\mathcal{A}}X,Z]\rightarrow[\Sigma^k X, Z]$ is bijective.

On the other hand one has the $\mathcal{A}$-cellularization $Cell_{\mathcal{A}}X\rightarrow X$, which is characterized as an $\mathcal{A}$-equivalence (i.e. it induces bijections on $[\Sigma^k A, \_]$ for all $A \in \mathcal{A}$ and $k \geq 0$) s.t. $Cell_{\mathcal{A}}X \in \mathcal{A}_+$.

If we take for $\mathcal{A}$ the slice cells with dimension $\geq n$ then it is claimed in [HHR][1] that $Cell_{\mathcal{A}}X$ is the fiber of the $\mathcal{A}$-nullification. However it seems that this is not formal and uses something about slice cells. Equivalently one can ask if the cofiber of the cellularization is the nullification. The second condition for nullifications is then automatically satisfied but the cofiber will only be $\mathcal{A}$-null if the map $[A, \Sigma Cell_{\mathcal{A}}X]\rightarrow [A, \Sigma X]$ is injective for all $A\in \mathcal{A}$. In the non-equivariant case this is true because for $\mathcal{A}={S^n}$ the cellularization $Cell_{\mathcal{A}}X$ is $(n-1)$-connected, so the map is injective for the trivial reason that the source vanishes. In the equivariant i see no reason why this argument should work because there will be non-trivial maps from slice cells in $\mathcal{A}$ to suspensions of slice cells in $\mathcal{A}$. Maybe someone can give me a hint as to why the map is injective or tell me if i'm overlooking a formal reason why the original statement is true. [1]: http://arxiv.org/pdf/0908.3724v3