Image in a locally finite abelian category over a field Let $K$ be a field. Let $\mathsf{C}$ be an abelian category such that for any two objects $X,Y \in \mathsf{C}$ the set $\operatorname{Hom}(X,Y)$ is a finite-dimensional $K$-vector space. Assume that every object $X \in \mathsf{C}$ has finite length.
Let $X,Y \in \mathsf{C}$. Let $\phi \in \operatorname{Hom}(X,Y)$ and $\alpha \in K$.
Is it true that the following holds?
\begin{equation}\operatorname{im}(\alpha \cdot \phi)\cong 
\begin{cases} \operatorname{im}(\phi) & \alpha \neq 0\\\ 
0 & \alpha=0
\end{cases}
\end{equation}
 A: I assume that $\mathsf{C}$ is actually supposed to be a $$-linear category.¹
If $α = 0$ then $α ϕ = 0$ because the zero element of $\mathrm{Hom}(X, Y)$ is the zero morphism.
The image of $α ϕ$ is then the image of the zero morphism, which is the zero subobject on $Y$.
So $\mathrm{im}(α ϕ) = 0$.
Suppose now that $α$ is nonzero.
The image of a morphism is the kernel of its cokernel.
It therefore suffices to show that $α ϕ$ and $ϕ$ have the same cokernel (in the sense that a morphism $Y \to Z$ is a cokernel of $α ϕ$ if and only of it is a cokernel of $ϕ$).
This is equivalent to asking that for every morphism $ψ \colon Y \to Z$ in $\mathsf{C}$ we have $ψ ∘ ϕ = 0$ if and only if $ψ ∘ (α ϕ) = 0$.
This equivalence is true, because we have $ψ ∘ (α ϕ) = α (ψ ∘ ϕ)$ by the $$-bilinearity of composition, and $α (ψ ∘ ϕ) = 0$ if and only if $ψ ∘ ϕ = 0$ because the scalar $α$ is nonzero.

¹ Not only do we have a $$-vector space structure on each $\mathrm{Hom}$-set, but the composition maps $\mathrm{Hom}(X, Y) × \mathrm{Hom}(Y, Z) \to \mathrm{Hom}(X, Z)$ are also $$-bilinear.
(This is, for example, the case if $\mathsf{C}$ is the category of modules over a $$-algebra.)
The addition of morphisms in $\mathrm{Hom}(X, Y)$ coming from the abelian structure of $\mathsf{C}$ does then agree with the addition that is part of the $$-vector space structure of $\mathrm{Hom}(X, Y)$.
(We have two (a priori) different preadditive structures on $\mathsf{C}$: one coming from the abelian structure of $\mathsf{C}$, and one coming from the $$-vector space structures on the $\mathrm{Hom}$-sets.
But on an additive category, there exists precisely one preadditive structure.
So both preadditive structures must agree.)
Consequently, the zero element of the vector space $\mathrm{Hom}(X, Y)$ is the zero morphism.
