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Let $\mathcal{A}$ be a semiadditive category (a category with all finite biproducts). Explicitely, $\mathcal{A}$ is a category with a zero object such that for every pair of objects $A_1,A_2\in\mathcal{A}$ there exists a diagram $A_i\xrightarrow{\iota_i}A_1\oplus A_2\xrightarrow{\pi_j}A_j$ such that

  1. $(A_1\oplus A_2,\iota)$ is a coproduct of $A_1$ and $A_2$
  2. $(A_1\oplus A_2,\pi)$ is a product of $A_1$ and $A_2$
  3. $\pi_j\circ\iota_i=\delta_{ji}$

I'm reading through the proof that $\mathcal{A}$ admits an enrichment over the category of abelian monoids (i.e. there is an abelian monoid structure on hom-sets and composition is bilinear). To show this we need some notation. For $A\in\mathcal{A}$, let

  1. $\Delta_A:A\rightarrow A\oplus A$ be defined by $p_i\Delta_A=\mathbf{1}_A$
  2. $\nabla_A:A\oplus A\rightarrow A$ be defined by $\nabla_A\circ\iota_i=\mathbf{1}_A$
  3. $t_A:A\oplus A\rightarrow A\oplus A$ be defined by $\pi_1\circ t_A\circ\iota_1=\pi_2\circ t_A\circ\iota_2=0$ and $\pi_1\circ t_A\circ\iota_2=\pi_2\circ t_A\circ\iota_1=\mathbf{1}_A$

When $\mathcal{A}$ is additive we can view these morphisms as matrices in the usual way and composition is given by matrix multiplication. For instance, $t_A= \begin{pmatrix} 0 & \mathbf{1}_A \\ \mathbf{1}_A & 0 \end{pmatrix}$ and $\Delta_A=\begin{pmatrix}\mathbf{1}_A\\ \mathbf{1}_A \end{pmatrix}$ so $t_A\circ\Delta_A=\begin{pmatrix} 0 & \mathbf{1}_A \\ \mathbf{1}_A & 0 \end{pmatrix}\begin{pmatrix}\mathbf{1}_A\\ \mathbf{1}_A \end{pmatrix}=\begin{pmatrix}\mathbf{1}_A\\ \mathbf{1}_A \end{pmatrix}=\Delta_A$.

In our situation, however, we do not have an additive structure (indeed, we're trying to define one!), so we need to be more careful when looking at these compositions.

So, I'm trying to prove that $t_A\circ\Delta_A=\Delta_A$ formally (just using the properties of products and coproducts) but I'm stuck.

If anyone's interested, the monoidal structure on $\mathcal{A}$ is defined as follows: for $f,g\in\mathrm{Hom}_{\,\mathcal{A}}(A_1,A_2)$ we define $f+g=\nabla_{A_2}\circ f\oplus g\circ\Delta_{A_1}$ where $f\oplus g:A_1\oplus A_2\rightarrow A_1\oplus A_2$ is defined by $\pi_1\circ f\oplus g\circ\iota_1=f$, $\pi_2\circ f\oplus g\circ\iota_2=g$, and $\pi_1\circ f\oplus g\circ\iota_2=\pi_2\circ f\oplus g\circ\iota_1=0$.

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Doesn't the word "biproduct" also include (or imply) a relation of the form $1_{A \oplus B} = \iota_1 \pi_1 + \iota_2 \pi_2$? I do not know how you want to include this in your formalism or whether it follows from the hypotheses, but it seems to be the key for calculating that $\pi_i (t_A \Delta_A) = \pi_i \Delta_A$ by inserting $1_{A\oplus A}$ between $t_A$ and $\Delta_A$, viz. $\pi_i (t_A \Delta_A) = \pi_i t_A 1_{A \oplus A} \Delta_A = \cdots = \pi_i \Delta_A$. –  Martin Apr 19 '13 at 7:39
    
The equation $\mathbf{1}_{A\oplus B}=\iota_1\pi_1+\iota_2\pi_2$ is supposed to be a consequence of the axioms given (see Pareigis section 4.1). Indeed, the definition of a biproduct is independent of a monoidal structure (see nlab entry for "biproduct"). However, maybe one could prove this equation first? using the definition of addition in the post we would want $\mathbf{1}_{A\oplus B}=\nabla_{A\oplus B}\circ\iota_1\pi_1\oplus\iota_2\pi_2\circ\Delta_{A\oplus B}$ but I'm not seeing why this is true either. –  Brian Fitzpatrick Apr 19 '13 at 14:21
    
It's a little too late at night for me to type out details, but $t_A$ has a universal property that can be stated entirely in terms of products (it's the braiding in the canonical symmetric monoidal structure on a category with finite products) and the conclusion follows pretty readily from here. –  Qiaochu Yuan Apr 21 '13 at 9:45
    
I'm having trouble figuring out the universal property of $t_A$. Any hints? –  Brian Fitzpatrick Apr 22 '13 at 0:06

1 Answer 1

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In any category with finite products there is a canonical map $\gamma_{A, B} : A \times B \to B \times A$ called the braiding. One might call it $(a, b) \mapsto (b, a)$, and indeed this gives a well-defined natural transformation of functors $\text{Hom}(-, A \times B) \to \text{Hom}(-, B \times A)$ giving rise to $\gamma_{A, B}$ via the Yoneda lemma. I claim that $t_A = \gamma_{A, A}$ and that this solves your problem because the definition of $\Delta_A$ is symmetric with respect to the two copies of $A$.

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