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Let $\mathsf{C}$ be a rigid abelian tensor category in the sense of Deligne and Milne's notes (p.9). Let $\mathbf{1}$ denote the identity object in $\mathsf{C}$ with respect to $\otimes$. The abelian group $\text{End}(\mathbf{1})$ has the structure of a commutative ring (see discussion immediately after Definition $1.15$). Let us suppose that this ring is actually a field, so by Remark $1.18$, $\mathbf{1}$ is a simple object. The proof of Proposition $1.19$ says that there is a criterion in $\mathsf{C}$:

$$X\neq 0 \iff X\otimes X^\vee \to \mathbf{1} \text{ is an epimorphism}.$$

Here the morphism $X\otimes X^\vee \to \mathbf{1}$ is the morphism obtained from $\operatorname{id}_{X^\vee}$ via the commutativity constraint $X\otimes X^\vee\cong X^\vee\otimes X$ and the bijection

$$\text{Hom}(X^\vee \otimes X, \mathbf{1})\cong\text{Hom}(X^\vee, X^\vee).$$

I have shown that if this morphism is an epimorphism then $X\neq 0$, but I'm having trouble working through the other implication (and I'm fairly sure this "criterion" wasn't stated previously). Could anyone show me how to prove that if $X\neq 0$ then the morphism is epic? Many thanks!


Edit: Now suppose $(\mathsf{C}', \mathbf{1}')$ is another rigid abelian tensor category with $\mathbf{1}\neq 0$ and let $F: \mathsf{C}\to\mathsf{C}'$ be an exact tensor functor. I must confess that even assuming the above criterion I am slightly mystified as to how this can be used to prove Proposition $1.19$:

$F$ is faithful

I understand that $F$ preserves the criterion, but I'm not sure how to relate this to showing it is injective on hom-sets between objects. Any tips on this would be greatly appreciated!

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For an exact functor $F$ between abelian categories, being faithful is equivalent to the condition $$FX=0\Rightarrow X=0$$ on objects, since a morphism is zero if and only if its image is zero, and exact functors preserve images.

So, for your second question, the "criterion" proves this equivalent formulation of faithfulness.

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    $\begingroup$ This answer, combined with working through the details of the proposition, has taught me a lot of basic but useful things about abelian categories and it is nice to see how the tensor/abelian structures interact. Thank you! $\endgroup$ – Alex Saad Apr 24 '16 at 10:29
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I think this answers the first part of my question (before the edit):

Suppose that the map $ev:X\otimes X^\vee \to \mathbf{1}$ is not an epimorphism. Then there exists an object $Z$ and distinct morphisms $\alpha, \beta: \mathbf{1}\to Z$ such that $\alpha\circ ev = \beta\circ ev$. Thus $ev: X\otimes X^\vee \to \mathbf{1}$ factors through the monomorphism $\ker(\alpha-\beta)\hookrightarrow \mathbf{1}$. Since $\text{End}(\mathbf{1})$ is a field, $\mathbf{1}$ is a simple object and therefore $\ker(\alpha-\beta)$ is isomorphic to either $\mathbf{1}$ or $0$ (as subobjects of $\mathbf{1}$ correspond to idempotents in $\text{End}(\mathbf{1})$). But if $\ker(\alpha-\beta) \cong \mathbf{1}$ then $\alpha-\beta$ is the zero map, which cannot happen since $\alpha\neq\beta$.

Hence $\ker(\alpha-\beta) = 0$ and so $ev: X\otimes X^\vee \to\mathbf{1}$ factors through the zero map $0\to\mathbf{1}$. By the tensor-hom adjunction, $\text{Hom}(X\otimes X^\vee, \mathbf{1})\cong \text{Hom}(X^\vee, X^\vee)$ is an isomorphism of abelian groups. Recall that $ev$ was defined as the image of $\text{id}\in\text{Hom}(X^\vee, X^\vee)$ under this adjunction, which implies $id_{X^\vee} = 0$. But then $X^\vee = 0$, so $X \cong (X^\vee)^\vee = 0$.

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