# Equality of Hom groups via 5 lemma

Reading a paper I found the following statement: given two spectra $A, B$ since multiplication by $p$ induces the same endomorphism in $[A,B]$ we have $[A \wedge M, B]\cong [A, \Sigma^{-1}M \wedge B]$, where $M$ is the mod $p$ Moore spectrum.

I think the idea is simple: since the smash with $M$ is just taking the cone of the multiplication by $p$ we have exact triangles in the stable homotopy category $A \xrightarrow{p} A \rightarrow A \wedge M$ and $\Sigma^{-1}M \wedge B \rightarrow B \xrightarrow{p} B$. Thus applying respectively the functors $[-, B]$ and $[A,-]$ we get two long exact sequences in the form

$[A,B]_{*+1} \xrightarrow{p} [A,B]_{*+1} \rightarrow Z \rightarrow [A,B]_{*} \xrightarrow{p}[A,B]_{*}$

where $Z$ is $[A \wedge M, B]_*$ or $[A, \Sigma^{-1}M \wedge B]_*$, so we should deduce that these two groups are isomorphic via the 5 lemma.

The point of my question is that there is no canonical map between them, so I do not know if the 5 lemma can be applied. We have two maps $[A \wedge M, \Sigma^{-1} M \wedge B] \rightarrow [A \wedge M, B]$ and $[A \wedge M, B] \rightarrow [A,B]$ but I do not understand if I can obtain a map making the diagram of long exact sequences commute.

The other option I see is that we can produce a map between the two groups via diagram chase using the fact that the other morphisms are invertible: I tried to do this but I cannot conclude the result. Since this kind of proof is immediate I suppose that this claim is false: it would be like proving that in the 5 lemma the vertical map in the middle is not needed.

Thanks in advance for any help.

I don't know the field you are talking about, but generally you need the central arrow. This corresponds to the fact that the sequence not always split: you can have non trivial "twist" of Ker and coker. For example, take $(\mathbb{Z}/p\mathbb{Z})^2, \mathbb{Z}/p^2 \mathbb{Z}$: they both fits into an exact sequence with Ker and coker equal to $\mathbb{Z}/p\mathbb{Z}$.
Another, more interesting, example. Consider central extensions by $h$ of a lie algebra $g$. These are exactly lie algebras which fits into an exact sequence with $h,g$, and are classified by $H^2(g,h)$.
A special case in which holds is the following: if you have $0 \to A \to B \to C \to 0$ exact, and $Ext^1(C,A)=0$, then every other $B'$ which fits the same exact sequence is isomorphic to $B$. This is a bit silly, because the condition I gave you allow to deduce that both $B,B'$ are isomorphic to $A \oplus C$.
I wonder if the result holds even in the following condition: if the map $Hom(C,C) \to Ext^1(C,A)$ coincide in the two exact sequences, then $B\simeq B'$ (this is trivially true if the Ext is zero).