# $Q$ is an injective object in an abelian category iff $\text{Hom}_C(\cdot, Q)$ is exact, the $\Rightarrow$ direction.

In an abelian category, I understand that $$\text{Hom}_C(X,Y)$$ forms an abelian group for all $$X, Y \in \text{Ob}(C)$$. Thus to show that $$Q$$ is an injective object implies $$\text{Hom}_C(\cdot, Q)$$ is an exact functor, we take the hom-sets to be objects in the category of abelian groups. Thus taking the $$\text{Hom}_C(0, Q) = 0$$ itself.

Thus assume $$Q$$ is an injective object, i.e. that for every monomorphism $$f : X \hookrightarrow Y$$ and morphism $$g : X \to Q$$ there is a morphism $$h : Y \to Q$$ such that $$h \circ f = g$$.

We want to show that the hom functor $$\text{Hom}_C(\cdot, Q)$$ applied to a short exact sequence in $$C$$:

$$0 \to X \xrightarrow{f} Y \xrightarrow{g} Z \to 0$$

yields another short exact sequence:

$$0 \leftarrow \text{Hom}_C(X, Q) \xleftarrow{- \circ f} \text{Hom}_C(Y, Q) \xleftarrow{-\circ g} \text{Hom}_C(Z, Q) \leftarrow 0$$ where $$- \circ f \equiv \text{Hom}_C(f, Q)$$ is notation.

So to show that the first on the left is exact we need to show that $$-\circ f$$ is surjective. I.e. for all $$g' : X \to Q$$ there exists a hom $$h : Y \to Q$$ such that $$(-\circ f) \circ h = g'$$. But that is true by definition of injective.

Next is to show exactness of the abelian group on the far right. And that is the same is showing that $$-\circ g$$ is injective. That is if $$g_1, g_2 \in \text{Hom}_C(Z, Q)$$ then $$g_1 \circ g = g_2 \circ g \implies g_1 = g_2$$.

Not sure how to prove that since there's nothing about uniqueness in the definition of injective object and the existing morphism $$h$$.

• Could you please me how to show the exactness in $\text{Hom}_C (Y,Q)$?
– Ryze
Commented Jun 15, 2020 at 11:49

You know that $$g$$ is an epimorphism, by the exactness of $$0 \to X\to Y \xrightarrow{g} Z\to 0$$. Hence the implication "if $$g_1 \circ g=g_2 \circ g$$ then $$g_1=g_2$$" is true for free, without using the injectivity of $$Q$$.
Indeed, for every object $$Q$$ it is the case that an exact sequence $$X \to Y \to Z \to 0$$ is sent by the functor $$\text{Hom}(-,Q)$$ to an exact sequence $$\text{Hom}(X,Q) \leftarrow \text{Hom}(Y,Q) \leftarrow \text{Hom}(Z,Q)\leftarrow 0$$, that is, $$\text{Hom}(-,Q)$$ is always left exact, even if $$Q$$ is not required to be injective. The injectivity of $$Q$$ is only used to prove that $$\text{Hom}(-,Q)$$ sends monomorphisms to surjections, as you did when you proved that $$\text{Hom}(f,Q)$$ is surjective by using that $$f$$ is monic and by the injectivity of $$Q$$.
• Okay, so I should just first flat out prove that $\text{Hom}(-, Q)$ is left exact. That's a big help! Thanks Commented Oct 3, 2019 at 20:17
• Yes. And observe indeed that we do not even use the injectivity of $Q$ in the proof of the exactness in $\text{Hom}(Y,Q)$. Commented Oct 3, 2019 at 20:19