Does the kernel of $f:\operatorname{End}_A\rightarrow\mathbb{Z}$ contains all non-automorphisms? Let:

*

*$A$ is a non-trivial abelian group

*$\operatorname{End}_A$ is the ring of endomorphisms of $A$

*$\operatorname{Aut}_A$ is the set of automorphisms of $A$

*$f:\operatorname{End}_A\rightarrow\mathbb{Z}$ is a unital ring homomorphism

*$\ker(f)$ is the kernel of $f$
We can prove that $\operatorname{Aut}_A\cap \ker(f)=\emptyset$. Does the same hold for the complements, i.e. $\overline{\operatorname{Aut}_A}\cap\overline{\ker(f)}=\emptyset$? Can we at least say that $\phi\notin\ker(f)$ implies that $\phi$ has the left or right inverse?
 A: The other answers are good, but I want to point out that this actually never happens.
Suppose for contradiction that $f : \operatorname{End}_A \to \mathbb{Z}$ such that $\ker(f)$ contains all non-automorphisms. Since $\ker(f)$ is a proper ideal of $\operatorname{End}_A$, it does not contain any elements of $\operatorname{Aut}_A$. This means that $\ker(f) = \operatorname{End}_A \setminus \operatorname{Aut}_A$. The existence of $f$ tells us that $\operatorname{End}_A \neq 0$, so we conclude $\operatorname{End}_A$ is a local ring with maximal ideal $\ker(f)$. Now $\operatorname{img}(f) \cong \operatorname{End}_A/\ker(f)$ must be a division ring. However, $\mathbb{Z}$ has no division subrings (these would be fields of characteristic $0$, and $\mathbb{Q}$ does not embed in $\mathbb{Z}$).
A: Let $A = \mathbb Z$ and $\chi: \mathbb Z \to \mathbb Z, \chi(z) = 2z$. Set $f: \operatorname{End}(A) \to \mathbb Z, f(\phi) = \phi(1)$. Then $f$ is a ring homomorphism because $f(\phi \circ \psi) = \phi \circ \psi(1) = \phi(\psi(1)) = \phi(n) = mn$, where $\phi(z) = mz$ and $\psi(z) = nz$. Moreover, $\chi \notin \ker f$.
A: I assume you prescribe that a ring morphism sends the unit to the unit, so that the map is surjective. This implies in particular that $\ker(f) $ is a proper ideal: if it was all of $End_A$, the image would be $0$.
The answer to the first question is yes: the automorphisms are by definition the units of the ring, and they can't intersect a proper ideal.
The answer to the second question is no as it is shown by the following example. Let $A=\mathbb{Z}$ and $ f: End_A \to \mathbb{Z} $ be defined by
$$f(\varphi) = \varphi(1) $$
Then the endomorphisms $\mu_n(x) =nx $ for $n\ge 2$ are not automorphisms and they are mapped to $n$ by $f$ (so they don't belong to the kernel) .
A: If $A=\mathbb{Z}$, then $\operatorname{End}(A)=\mathbb{Z}$. You can then take $f$ to be the identity of $\mathbb{Z}$, and any element which is neither $0$ nor $\pm 1$ is neither in the kernel nor invertible.
