# Uniqueness of morphisms that make sequence exact (hom functor)

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 $$j \in \operatorname{Hom}(A,A)$$?

• Isn’t this the definition of the $\operatorname{Hom}$ function though? It’s not about coming how with maps that make the exact sequence work, but checking the maps which are given by the hom function make the exact sequence work Commented Aug 14 at 1:55
• Yes, I understand the proof that f’ and g’ given by the definition of hom functor makes the exact sequence work. What I want to know is, regardless of it, whether the maps making sequence exact are unique.
– 이상원
Commented Aug 14 at 2:00
• Unique in what sense? Like the only maps that make this a short exact sequence? Commented Aug 14 at 3:12
• Yes, that's what I've meant
– 이상원
Commented Aug 14 at 3:32

## 1 Answer

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.

• Thanks. Is it true that $h'$ is unique?
– 이상원
Commented Aug 14 at 3:35
• No, for example, if $h’$ makes the sequence$$0\to\operatorname{Hom}(A,A)\xrightarrow{h’}\operatorname{Hom}(A,E)\to\operatorname{Hom}(A,F)$$ exact, then $-h’$ also does.
– Piyo
Commented Aug 14 at 6:42