Uniqueness of morphisms that make sequence exact (hom functor)

Question

Let $\mathcal{A}$ be abelian category, and $X\in\operatorname{ob}(\mathcal{A}).$ $\operatorname{Hom}(X,-)$ is left exact functor from $\mathcal{A}$ to $\mathsf{Ab}$, which means that when $0 \to A \xrightarrow[]{f} B \xrightarrow[]{g} C$ is exact sequence in $\mathcal{A}$, $0 \to \operatorname{Hom}(X,A) \xrightarrow[]{f'} \operatorname{Hom}(X,B) \xrightarrow[]{g'} \operatorname{Hom}(X,C)$ is exact in $\mathsf{Ab}$, given $f'$ by $h\mapsto f\circ h$ for all $h\in \operatorname{Hom}(X,A)$ and $g'$ by $i\mapsto g\circ i$ for all $i\in \operatorname{Hom}(X,B)$.

I understand that it is sufficient to show that there exists, at least one pair of $f'$ and $g'$ with properties of exact sequences($\operatorname{ker} f'=0, \operatorname{im}f'=\operatorname{ker}g'$)exists to prove that $0 \to \operatorname{Hom}(X,A) \xrightarrow[]{f'} \operatorname{Hom}(X,B) \xrightarrow[]{g'} \operatorname{Hom}(X,C)$ is exact. Is pair of $f'$ and $g'$ given by $h\mapsto f\circ h$ and $i\mapsto g\circ i$ unique solution?

This question arised from reading the proof of this proposition:

If $E \xrightarrow[]{f} F$ is morphism in $\mathcal{A}$ with kernel $\alpha: A\to E$, then $f \circ \alpha=0$

Proof: Using definition of kernel by Serge Lang, $0 \rightarrow \operatorname{Hom}(A,A)\xrightarrow[]{h'} \operatorname{Hom}(A,E) \rightarrow \operatorname{Hom}(A,F)$ is exact. $\alpha$ is image of ${id}_A$ by $h'$, so it is in the kernel of second morphism.

Can $h'$ only be $j \mapsto \alpha\circ j$ for all $j \in \operatorname{Hom}(A,A)$?

Answer 1

No, the maps $f’$ and $g’$ that make the sequence $$ 0\to\operatorname{Hom}(X,A)\xrightarrow{f’}\operatorname{Hom}(X,B)\xrightarrow{g’}\operatorname{Hom}(X,C) $$ exact are not unique. In fact, if $f’$, $g’$ are such maps, then $-f’$, $g’$ are also.