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I'm using the following definition:

A category $\mathcal C$ is given by the following:

  1. A collection of objects $A$
  2. A set $M(A, B)$ for any two objects $A, B \in \mathcal C$.
  3. A function $M(B, C) \times M(A, B) \to M(A, C)$ for each triple of objects $A, B, C \in \mathcal C$.

These terms must satisfy the following axioms

I: If $\alpha : A \to B$, $\beta : B \to C$ and $\gamma : C \to D$, then $\gamma(\beta \alpha) = (\gamma \beta)\alpha$.

II: For each $A \in \mathcal C$, there exists $i_A : A \to A$ such that if $\beta : B \to A$, $\gamma : A \to C$ then $i_A \beta = \beta$ and $\gamma i_A = \gamma$.

A category $\mathcal C$ shall be called a $K$-category where $K$ is a commutative ring with unity if it has a distinguished object 0, the zero object, and satisfies the following three axioms:

III: $M(A, B)$ is a unital $K$-module.

IV: The function $M(B, C) \times M(A, B) \to M(A, C)$ is a bilinear map of $K$-modules

V: $M(0, 0)$ is the zero module, also written 0.

The following theorem is what I am having trouble with. I can't seem to be able to get the result using the previous definition of $K$-category (From Swan's 'The Theory of Sheaves')

Prove: $M(0, A) = M(0, 0) = M(A, 0)$.

Proof:Since we know $M(A, B)$ is a set (moreover module) for any $A, B \in \mathcal C$, then it would suffice to show that $M(0, A) \subseteq M(0, 0)$ and $M(0, 0) \subseteq M(0, A)$. The latter inclusion is trivial since $M(0, 0)$ is the zero module and every module has the zero module as a submodule.

Now let $f \in M(0, A)$. We want to show that $f \in M(0, 0)$, but since $M(0, 0)$ is the zero module and it is an additive subgroup, its only element is the 'zero'-map such that the following two axioms hold: (Axioms of module)

  1. $f + 0 = f$ for all $f \in M(A, B)$
  2. $f + (-f) = 0$ for all $f \in M(A, B)$

Here is where I am stuck. I can't seem to get the result directly from using the axioms I was given. Any help?

As a sub-question (soft), does anyone have any opinions/recommendations on my choice of textbook on Sheaf Theory? This is my first introduction to the subject and I wanted a textbook that didn't have too much topology and this seemed great, but I've been finding a bit of types :(

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1 Answer

up vote 2 down vote accepted

Let $f\in M(0,A)$. The axiom IV means in particular that $(f\cdot k)i_0=f(k\cdot i_0)$ for $k\in K$. Take $k=0$, then $0_A=fi_0$, i.e. $f=0_A$.

Remark: It is incorrect to write $M(0, A) \subseteq M(0, 0)$ since $M(A,B)\cap M(C,D)=\emptyset$ for $(A,B)\ne (C,D)$.

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Can you confirm: I understand how you use bilinearity. If you take $k=0$, then $f \cdot 0 = 0_A$ where $0_A$ is the zero element of the $M(0, A)$ module (property of modules, scalar times zero = zero of module), so $(f \cdot k)i_0 = 0_A i_0 = 0_A$. But then $(0 \cdot i_0) = 0_0 = i_0$ because $M(0, 0)$ only has one element, so $f(0 \cdot i_0) = f i_0 = f$, correct, showing $f = 0_A$? –  Robert Sep 19 '13 at 13:34
    
More importantly, if I am not trying to show a double inclusion, then what am I trying to show? An isomorphism of modules? The both only have one element so that 'are' both the zero module, up to isomorphism? Thanks a lot :D –  Robert Sep 19 '13 at 13:36
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To 1st remark: Yes, you understand right. To 2nd remark: Yes, in both cases one should prove triviality of the modules. –  Boris Novikov Sep 19 '13 at 14:10
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