Proving that $\mathrm {Hom}(X,-)$ is left exact in abelian categories I am following Pavel et al's book "Tensor categories".
They claim without proof that the (covariant) functor $F:=\mathrm {Hom}(X,-):\mathcal C \rightarrow \textbf{Ab}$ is left exact, where $\mathcal C$ is an abelian category. I am trying to show this. I am new to this topic.
Take $0\xrightarrow{a} A\xrightarrow{b} B\xrightarrow{c} C\rightarrow 0$ short exact. We want to show that
 $$0\rightarrow FA\rightarrow FB\rightarrow FC$$
 is exact.
I started by trying to show that im$Fa = \ker Fb$. since im$Fa=\ker \mathrm {coker} Fa$, then $(\mathrm {coker}Fa) \circ (\mathrm{im} Fa)=0$.
I was trying to use this to show that $0\xrightarrow{\mathrm{im}Fa}FA \underset{0}{\overset{Fb}{\rightrightarrows}} FB$ is an equalizer diagram and thus complete the proof. But I got nowhere. 
 A: Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$$
be short exact. We want to show that 
$$0 \to \hom(X,A) \xrightarrow{\iota \circ }  \hom(X,B) \xrightarrow{\pi\circ} \hom(X,C) $$
 is exact. where the morphisms are just the compositions with $\iota$ and $\pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $\iota \circ g = \iota \circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE. 
Exactness at $\hom(X,A)$: 
Here it suffices to show that $\iota \circ$ is injective, but this just means that $\iota \circ g =\iota \circ g'$ implies that $g=g'$, which holds since $\iota$ is monic.
Exactness at $\hom(X,B)$:
We will prove $\mathrm{im}(\circ\iota) \subset \ker(\circ \pi)$ and  $\ker(\circ \pi) \subset \mathrm{im}(\circ\iota) $ seperately.
$\mathrm{im}(\circ\iota) \subset \ker(\circ \pi)$: 
Let $f \in \mathrm{im}(\circ\iota)$ this means that there is a $g$ such that $f=\iota \circ g$, but this now gives $$\pi \circ f = \pi \circ \iota \circ g = 0 \circ g =0$$.
$\ker(\circ \pi) \subset \mathrm{im}(\circ\iota) $
Let $f \in \ker(\circ \pi)$, then we have $\pi \circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $\iota$ actually define a kernel of $\pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A \xrightarrow{\iota} B$, but this literally means that there is a $g$ s.t $f= g\circ \iota$ which finishes the claim.
