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I am reading an article and it is written there: If A is a k-linear category (possibly without direct sums) we can embed it in the additive category A × N, where a morphism (x,m) → (y, n) is an n × m matrix with entries in A(x, y) = HomA(x, y). Of course if A is additive then A ≈ A × N. Could anyone explain it? Why is it additive?

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up vote 5 down vote accepted

It looks like if this is a wrong construction of the additivization of a $k$-linear category. Either this, or I don't understand the notation. Let me just sketch how the additivization $\operatorname{Add}(\mathcal{A})$ of a $k$-linear category $\mathcal{A}$ can be constructed:

Objects in $\operatorname{Add}(\mathcal{A})$ are finite sequences $(X_1,\dots, X_n)$ of objects in $\mathcal{A}$, $n\geq 0$. Morphism $k$-modules are as follows:

$$\operatorname{Add}(\mathcal{A})((X_1,\dots, X_n),(Y_1,\dots, Y_m))= \bigoplus_{i=1}^n\bigoplus_{j=1}^m\mathcal{A}(X_i,Y_j). $$ In particular, a morphism $\varphi\colon (X_1,\dots, X_n)\rightarrow(Y_1,\dots, Y_m)$ is a matrix $\varphi=(\varphi_{ij})$ formed by morphisms $\varphi_{ij}\colon X_i\rightarrow Y_j$. Composition is defined by matrix multiplication and composition in $\mathcal{A}$. Direct sums can be calculated as $$(X_1,\dots, X_n)\oplus(Y_1,\dots, Y_m)=(X_1,\dots, X_n,Y_1,\dots, Y_m).$$ The empty sequence $()$ is a zero object.

The $k$-linear functor $$ \mathcal{A}\longrightarrow\operatorname{Add}(\mathcal{A}),$$ $$X\mapsto(X),$$ is fully faithful and universal (initial) among $k$-linear functors from $\mathcal{A}$ to $k$-linear additive categories.

If $\mathcal{A}$ is additive, a choice of direct sums defines an inverse equivalence, $$ \operatorname{Add}(\mathcal{A})\longrightarrow \mathcal{A}, $$ $$(X_1,\dots, X_n)\mapsto X_1\oplus\cdots\oplus X_n.$$

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