# In a pretriangulated category, a morphism is an isomorphism if and only if its homotopy kernel and homotopy cokernel are zero

Let $$\mathcal{T}$$ be a pretriangulated category with suspension $$\Sigma$$ (assumed to be an automorphism) and a class of distinguished triangles. $$v\colon Y\to Z$$ is a homotopy cokernel of $$u\colon X\to Y$$ if there is a distinguished triangle $$X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} \Sigma X$$. Dually, $$t\colon T\to X$$ is a homotopy kernel if there is a distinguished triangle $$X \xrightarrow{u} Y \xrightarrow{v} \Sigma T \xrightarrow{\Sigma t} \Sigma X$$. These notions are unique up to non-unique isomorphism. For the reference, I use Murfet's notes:

http://therisingsea.org/notes/TriangulatedCategories.pdf

Murfet attempts to prove that $$u$$ is an isomorphism if and only if its homotopy kernel and homotopy cokernel are zero. One direction is proven correctly. But in proving that a morphism with zero homotopy kernel and homotopy cokernel is an isomorphism, he seems to make an obvious mistake (page 9).

So, what should be the right way to prove that? One useful lemma Murfet proves beforehand is the following one:

$$u\colon X\to Y$$ is an isomorphism if and only if there is a distinguished triangle $$X \xrightarrow{u} Y \xrightarrow{} 0 \xrightarrow{} \Sigma X$$.

The problem is, a homotopy kernel and a homotopy cokernel of $$u$$ are zero, then there is a triangle $$X \xrightarrow{u} Y \xrightarrow{0} Z \xrightarrow{0} \Sigma X$$, but $$Z$$ is not necessarily zero.

Also, I would like to avoid the use of enhancements (but it's unlikely they are needed since the problem seems to be rather elementary).

## 1 Answer

It follows from the second to next result in the notes, namely Proposition 12: a morphism with zero homotopy kernel (resp., homotopy cokernel) is necessarily a split monomorphism (resp., split epimorphism). Of course, a morphism which is a split monomorphism and a split epimorphism is necessarily an isomorphism.