Krdytkyu

I am trying to solve the following exercise.

Let $mathcal A$ be an Abelian category and consider the category of cochain complexes in $mathcal A$.

Construct an example of a cochain complex $(A^bullet, d^bullet)$ whose cohomology is concentrated in a single degree and which does not admit a non-zero chain map from (n)or to its cohomology.

Constructing an example of a complex which does not admits a chain map from its cohomology is straight-forward (e.g. $mathbb Z overset{times 2}{longrightarrow} mathbb Z$).

I'm hoping to construct an example for the "to" part with $mathcal A$ being the category of finitely generated Abelian groups, but I don't see how. Needless to say, I also don't know how to do the "neither from nor to" part.

This is not possible if you restrict your attention to finitely generated abelian groups - there are no solutions to the "to" problem, let alone the "neither to nor from" problem. There is a proof below. Below that there is a hint about how to find an example in the category of all Abelian groups.

There are no examples in fg abelian groups. If $A$ is a complex, and $B$ is the complex with $H^i(A)$ in degree $i$, then $$Hom(A, B) = Hom_{Ab}(A^i, H^i(A))$$

So the question is equivalent to

Equivalent question. Does there exist a fg abelian group $A$ with nonzero subquotient $B = A'/A''$ ($A'' leq A' leq A$) such that $Hom(A,B) = 0$? The answer is no.

For non-zero finitely generated abelian groups $A,B$ we have $Hom(A,B) = 0$ only if $A$ is torsion. For instance, if we write $A = A_{free}oplus A_{tors}$ with $a = |A_{tors}|$ and similarly for $B$ and $b$ then we have:
$$Hom(A_{free}, B) = 0 implies A_{free} = 0,$$
$$Hom(A_{tors}, B_{tors}) = 0 implies (a,b) = 1,$$
$$Hom(A_{tors}, B_{free}) = 0 implies true$$
Since $B$ is a subquotient of $A$ it must also be torsion and $b|a$. But $b|a$ with $(a,b) = 1$ implies $a = b = 1$, i.e. $A = B = 0$.

Suggestions. To find a counter example in another category, you can still use the above "equivalent question". You need to find, say, a module $A$ with a subquotient $A'/A''$ and no maps $A to A'/A''$. This corresponds to the complex ${cal A} = [A'' to A to A/A']$ whose cohomology is just $H^1(mathcal{A}) = A'/A''$.

An even easier version of this problem is when $A'' = 0$. Then you are just looking for an object $A$ with a subobject $A'$ such that there are no maps $A to A'$. It's possible to find an example like this in Abelian groups (not finitely generated!). See if you can find it.

If you get stuck here is probably the simplest solution:

${cal A} = [mathbb Q to mathbb Q/mathbb Z]$.

Note: By picking a suitable $A''$, you can also extend this example to a "neither to nor from" example!