A spectrum $I$ is $E$-injective iff the map $i:I\rightarrow I\wedge E$ is an inclusion of a retract. I was reading some notes on stable homotopy theory and I came across the statement in the title of this question. 
"Suppose $E$ is a ring spectrum, then $I$ is $E$-injective if and only if the natural map $i:I\rightarrow I\wedge E$ is an inclusion of a retract."
I'm not sure how widely used this notation is, so I'll write down the definitions needed here (which reflect definitions from homological algebra).
$\bullet$ A sequence of maps $\cdots\rightarrow X\rightarrow Y\rightarrow Z\rightarrow \cdots$ of spectra is $E$-exact if after smashing with $E$ it induces an exact sequence of homotopy functors.
$$\cdots\rightarrow[-,X\wedge E]\rightarrow [-, Y\wedge E]\rightarrow[-,Z\wedge E]\rightarrow\cdots$$
$\bullet$ A map $f:X\rightarrow Y$ is an $E$-monomorphism if $\ast\rightarrow X\rightarrow Y$ is $E$-exact.
$\bullet$ A spectrum $I$ is $E$-injective if for each $E$-monomorphism $f:X\rightarrow Y$ and map $g:X\rightarrow I$, there exists a map $h:Y\rightarrow I$ such that $h\circ f=g$. In other words, $h$ extends $g$.
$\bullet$ A corollary of this information and the fact that $E$ is a ring spectrum tell us that the natural map in question $i:X\rightarrow X\wedge E$, which is defined as $id_X\wedge \eta:X\wedge S^0\rightarrow X\wedge E$ where $\eta:S^0\rightarrow E$ is the unit for $E$, is in fact an $E$-monomorphism for any spectrum $X$. This corollary comes almost straight from the fact that $E$ is a ring spectrum.
One direction of this proof is now simple: If $I$ is $E$-injective, then we have an $E$-monomorphism $i:I\rightarrow I\wedge E$ and a map $id_I:I\rightarrow I$, so using the fact that $I$ is $E$-injective we obtain a map $r:I\wedge E\rightarrow I$ such that $r\circ i=id_I$, so $i$ is an inclusion of a retraction.
It is the other direction that confuses me. Let $i:I\rightarrow I\wedge E$ be the inclusion of a retract $r:I\wedge E\rightarrow E$, so $r\circ i=id_I$. Now given an $E$-injection $f:X\rightarrow Y$, and a map $g:X\rightarrow I$, I would like to produce a map $h:X\rightarrow I$ but I'm finding this difficult!
If I could even find a map $\widetilde{h}:X\wedge E\rightarrow I\wedge E$ then I think I would be find because I could just precompose with my $E$-mono $i:X\rightarrow X\wedge E$ and postcompose with my retraction $r:I\wedge E\rightarrow I$. I think this is where I might use the hypothesis that such a retraction exists, but it still does not give me my nice map $h:X\rightarrow I$.
I really want both directions of this statement, because this means that the standard and normalised $E$-Adams resolutions are effectively injective resolutions analogous to homological algebra.
I would appreciate anyone's opinion on this!
 A: After doing some more research into the topic, I've noticed that the definition I've given above for an $E$-injective spectrum is not the definition everyone makes in a triangulated category setting. In some 2012 notes by Haynes Miller, the definition of an $E$-injective spectrum $I$ is exactly one where the canonical map $i:I\rightarrow I\wedge E$ is an inclusion of a retract.
Hence my question above is really asking for an equivalence of these two definitions. I believe these definitions are equivalent in any triangulated category with a coherent monoidal structure, but I'll just work with our stable homotopy category of orthogonal spectra for this.
Once direction of the proof is given in my question above. For the other direction, assume that we have a spectrum $I$ with maps $i:I\rightarrow I\wedge E$ and $r:I\wedge E\rightarrow I$ such that $r\circ i=id_I$. Suppose now that we have a map $f:X\rightarrow Y$ such that 
$$(f\wedge id_E)_\ast:[A,X\wedge E]\longrightarrow [A,Y\wedge E]$$
is a monomorphism for all spectra $A$, and another map $g:X\rightarrow I$. Let's take the cofibre of the map $f\wedge id_E$ to obtain a distinguished triangle
$$X\wedge E\longrightarrow Y\wedge E\longrightarrow C\longrightarrow \Sigma(X\wedge E),$$
and recall that $I\wedge E\rightarrow I\wedge E\rightarrow \ast\rightarrow \Sigma(I\wedge E)$ is a distinguished triangle. Since $(f\wedge id_E)_\ast$ is a 
monomorphism, then $\phi_\ast:[A,C]\rightarrow [A,\Sigma(X\wedge E)]$ is the zero map for all spectra $A$, since distinguished triangles induce long exact sequences on these represented functors. By the Yoneda Lemma, this tells us the map $\phi:C\rightarrow \Sigma(X\wedge E)$ which induces the natural transformation $\phi_\ast$, must be trivial. Hence we have the following diagram of distiguished triangles.
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{ccccccc}
X\wedge E & \ra{f\wedge id} & Y\wedge E & \ra{} & C & \ra{\phi} & \Sigma(X\wedge E)\\
\da{g\wedge id} & &  & & \da{} & & \da{\Sigma g} \\
I\wedge E & \ra{id} & I\wedge E & \ra{} & \ast & \ra{} & \Sigma(I\wedge E)\\
\end{array}
$$
By the triangulated category axioms, we have a map $\widetilde{h}:Y\wedge E\rightarrow I\wedge E$, such that $\widetilde{h}(f\wedge id_E)=g\wedge id_E$. In my question above I claim that setting $h=r_I\widetilde{h}i_Y$ will work, so let's quickly show this. First notice that the maps $i_A:A\rightarrow A\wedge E$ are natural in $A$, and that we have a map $r_I$ with $r_I i_I=id_I$ (a fact we haven't used yet!).
$$hf=r_I\widetilde{h}i_Yf=r_I\widetilde{h}(f\wedge id_E)i_X=r_I(g\wedge id_E)i_X=r_Ii_Ig=g$$
Phew. Now we have a map $h:Y\rightarrow I$ which helps factor $g:X\rightarrow I$ through this $E$-monomorphism $f:X\rightarrow Y$, so our two definitions of an $E$-injective spectrum are equivalent.
