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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.

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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.

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  • $\begingroup$ Very good answer. But I have a couple of questions: 1. You say that $\iota$ is a kernel of $\pi$, but it is im $\iota$ instead, no? 2. You use that $\ker c = \mathrm{im} b \Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = \ker \mathrm{coker} b$ but didn't get it. $\endgroup$ – Soap Nov 23 '18 at 15:52
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    $\begingroup$ No, $(\mathrm{im}(\iota), \mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,\iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel. $\endgroup$ – Enkidu Nov 26 '18 at 8:53
  • $\begingroup$ I understand that I can write $b=im(b) \circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel... $\endgroup$ – Soap Dec 1 '18 at 14:14
  • $\begingroup$ Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here $\endgroup$ – Enkidu Dec 3 '18 at 9:57
  • $\begingroup$ Why is $(A,l)$ a kernel of $\pi $? I can't point out why it is the case. $\endgroup$ – CTW Mar 14 at 15:46

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