# 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.

## 1 Answer

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).

• I am studying a way to classify "extension" via homological algebra; if you are interested in, I can write it down. Of course it is difficult to apply in practice because it relies on things like projective resolutions which are usually hard to compute. But maybe your context is sufficiently general. – frame95 Jun 15 '18 at 20:07