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I am currently trying to understand why finite products and coproducts in the category $\text{Ab}$ coincide. In fact, I'm not even sure I can show it. My question is the following:

Is there an intuitive way to understand why finite products and coproducts in $\text{Ab}$ coincide, while the same is not true in $\text{Grp}?$

As for the formal direction, I'm not sure if I have shown that the two coincide. The idea is to show that $G\times H$ satisfies the universal property for coproducts for $G,H \in \text{Obj}(\text{Ab})$.

Attempt

I defined the inclusion functions $\iota_G:G\longrightarrow G\times H$ and $\iota_H:H\longrightarrow G\times H$ by $\iota_G(g)=(g,1_G)$ and $\iota_H(h)=(1_H,h)$. Suppose $\varphi:G\longrightarrow K$ and $\psi:H\longrightarrow K$ are homomorphisms. To show $G\times H$ is a coproduct in $\text{Ab}$, I then need to construct a unique function $\tau:G\times H\longrightarrow K$ such that $\varphi=\tau\circ\iota_G$ and $\psi=\tau\circ\iota_H$. As such, I defined by $\tau$ by $\tau(g,1_H)=\varphi(g)$ and $\tau(1_G,h)=\psi(h)$. I think this should work since for all $(g,h) \in G\times H$ we have $(g,h)=(g,1_H)*(h,1_H)$, so by defining $\tau$ so that $\tau(g,h)=\tau(g,1_H)*\tau(h,1_H)$ everything should work.

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Note, only finite coproducts are isomorphic with finite products... –  Thomas Andrews Nov 24 '12 at 16:23
    
@thomasandrews thank you, I was not aware of that. I am, in this question, only interested in the coproduct of two groups, but I will edit to make this clear. –  Holdsworth88 Nov 24 '12 at 16:25
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It's a consequence of $\mathsf{Ab}$ being an additive category (in fact, an abelian category): in any additive category, by definition, products and coproducts coincide (and are called biproducts). If I get time then I'll flesh this out into a proper answer later. –  Clive Newstead Nov 24 '12 at 16:31
    
@clivenewstead I would be eternally grateful. –  Holdsworth88 Nov 24 '12 at 16:32
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@CliveNewstead: This is already true in a pre-additive category. If in a pre-additive category the product and coproduct of two objects both exist, they must coincide. An explicit isomorphism is given by the map $$in_1 \circ pr_1 + in_2 \circ pr_2: X \times Y \to X \sqcup Y,$$ where $in_1$ and $in_2$ are the inclusion maps and $pr_1$ and $pr_2$ are the projection maps of course. –  Alexander Thumm Nov 24 '12 at 16:47

4 Answers 4

up vote 9 down vote accepted

The category $\mathsf{Ab}$ of abelian groups and their homomorphisms is an example of a preadditive category. In a preadditive category, the notions of (finitary) product and coproduct coincide, i.e. $$A \times B \cong A + B \cong A \oplus B$$ where $\oplus$ denotes the biproduct of $A$ and $B$.

In an arbitrary category, the biproduct is defined as follows: $X$ is a biproduct of $A$ and $B$ if there exist maps $$A \overset{\pi_A}{\underset{\iota_A}{\leftrightarrows}} X \overset{\pi_B}{\underset{\iota_B}{\rightleftarrows}} B$$ making $(A \overset{\pi_A}{\leftarrow} X \overset{\pi_B}{\rightarrow} B)$ into a product and $(A \overset{\iota_A}{\rightarrow} X \overset{\iota_B}{\leftarrow} B)$ into a coproduct.

Now, preadditive categories have an additive structure on their homsets. For example, in $\mathsf{Ab}$, if $f, g : A \rightrightarrows B$ are homomorphisms between abelian groups $(A,+)$ and $(B,+)$ then we can obtain another homomorphism $f+g : A \to B$ defined by $$(f+g)(a)=f(a)+g(a)$$ Then the binary operation $+$ gives $\text{Hom}_{\mathsf{Ab}}(A,B)$ an abelian group structure. The identity element is the zero map $0 : A \to B$, which is a zero morphism in the category theoretic sense.

There is a result which states that $(A \overset{\pi_A}{\underset{\iota_A}{\leftrightarrows}} X \overset{\pi_B}{\underset{\iota_B}{\rightleftarrows}} B)$ is a biproduct if and only if all of the following are satisfied:

  • $\pi_A \circ \iota_A = \text{id}_A$ and $\pi_B \circ \iota_B = \text{id}_B$
  • $\pi_B \circ \iota_A = 0 : A \to B$ and $\pi_A \circ \iota_B = 0 : B \to A$
  • $\iota_A \circ \pi_A + \iota_B \circ \pi_B = \text{id}_X$

This hints at why it works in $\mathsf{Ab}$ but not in $\mathsf{Grp}$: in $\mathsf{Grp}$ we don't have additive structure on homsets like $\mathsf{Ab}$ does. [And, indeed, the coproduct of two groups in $\mathsf{Grp}$ is the free product; in $\mathsf{Ab}$ it is the direct sum $\equiv$ Cartesian product.]

It's worth trying to prove the above result and work through the proof with a (more) concrete example.


P.S. Thanks Alexander Thumm for reminding me about preadditive categories.

Edit: See comments.

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This is not the correct definition of a biproduct (and in particular the theorem you cite is not true with this definition). One reason is that it does not make the biproduct unique up to unique isomorphism like a universal construction ought to be. You need to impose additional conditions to get that uniqueness (namely you need to fix the four possible compositions of the structure maps); see, for example, qchu.wordpress.com/2012/09/14/… . –  Qiaochu Yuan Nov 25 '12 at 7:56
    
@QiaochuYuan I don't think so. There is a not-so-well-known theorem of Lack saying that in any pointed category, the existence of any natural isomorphism $X + Y \cong X \times Y$ is enough to ensure the canonical map $X + Y \to X \times Y$ is an isomorphism. –  Zhen Lin Nov 25 '12 at 8:19
    
@Zhen: the definition given by Clive doesn't posit a natural isomorphism, it posits an isomorphism. –  Qiaochu Yuan Nov 25 '12 at 8:28
    
There are some subtle points here. There is a natural isomorphism, but it isn't the identity transformation. Once you have a semiadditive structure, you can show that $\langle \textrm{id}_A, 0 \rangle : A \to A \times B$ and $\langle 0, \textrm{id}_B \rangle : B \to A \times B$ make $A \times B$ into a coproduct. Let's call it the canonical coproduct structure. Now, if $A \times B$ has any other coproduct structure $(A \times B, \iota_1, \iota_2)$ , we get a unique natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to the canonical coproduct structure on $A \times B$. –  Zhen Lin Nov 25 '12 at 9:06
    
Of course, the coproduct and product functors are literally equal provided we use the canonical coproduct structure, so the natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to the canonical coproduct structure is also a natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to $(A \times B, \pi_1, \pi_2)$. Obviously this reasoning means we can't apply Lack's theorem here, but nonetheless it suggests Clive's definition is at worst unwieldy, rather than wrong. –  Zhen Lin Nov 25 '12 at 9:12

As Thomas Andrews indicates, it might be illuminating to consider products and coproducts of possibly infinite collections of Abelian groups. Try to understand the following two facts:

1) Products and coproducts of Abelian groups correspond to direct products and direct sums, respectively.

2) For finite collections of Abelian groups, the direct product and the direct sum happen to be canonically isomorphic.

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Let $\mathcal A = \text{AbGrp}$. It will be important later that $\mathcal A$ has a zero object $0$ and zero morphisms.

The key property is that every hom-set $\mathcal A(A,B)$ is an abgroup. Let $f_1, f_2 : A \longrightarrow B$ then define $f_1 + f_2 : A \longrightarrow B$ simply by $(f_1 + f_2)(a) = f_1(a) + f_2(a)$, it's easy to see this is a group homomorphism. Furthermore, in $\mathcal A$, composition distributes: We have the identities $g(f_1+f_2) = gf_1 + gf_2$ and $(g_1+g_2)f = g_1f + g_2f$.

That doesn't happen in $\mathcal G = \text{Grp}$ suppose we let $(f_1 \star f_2)(a) = f_1(a) \star f_2(a)$, then $v$ isn't a group homomorphism because $(f_1 \star f_2)(a \cdot b) = f_1(a) \star f_1(b) \star f_2(a) \star f_2(b)$ is not in general equal to $(f_1 \star f_2)(a) \star (f_1 \star f_2)(b) = f_1(a) \star f_2(a) \star f_1(b) \star f_2(b)$.


Now let me introduce the biproduct $A \oplus B$ which is an object with maps

  • $\pi_1 : A \oplus B \longrightarrow A$
  • $\pi_2 : A \oplus B \longrightarrow B$
  • $\iota_1 : A \longrightarrow A \oplus B$
  • $\iota_2 : B \longrightarrow A \oplus B$

satisfying the identities

  • $\pi_1 \iota_1 = 1_A$
  • $\pi_2 \iota_2 = 1_B$
  • $\pi_2 \iota_1 = 0$
  • $\pi_1 \iota_2 = 0$
  • $\iota_1 \pi_1 + \iota_2 \pi_2 = 1_C$

notice that this definition is self dual.


Lemma $C$ is the product of $A$ and $B$ iff it is the* biproduct of $A$ and $B$. (we haven't show biproducts unique but they are).

proof sketch: ($\Rightarrow$) You can construct $\iota_1, \iota_2$ using the equations and you can prove the last identity by showing first $\pi_1 (\iota_1 \pi_1 + \iota_2 \pi_2) = pi_1$ then $\pi_2 (\iota_1 \pi_1 + \iota_2 \pi_2) = pi_2$ . ($\Leftarrow$) One just needs to show that the biproduct has the universal property of products (the uniqueness of a universal map).

Corollary (by duality) $C$ is the coproduct of $A$ and $B$ iff it is the* biproduct of $A$ and $B$.

Corollary Products and Coproducts coincide.


The properties of $\mathcal A$ we used are called a preadditive category but I tried make this as elementary as possible so just note that $\mathcal A$ is one example and $\mathcal G$ is not.

We have not yet seen why products and coproducts in $\mathcal G$ do not coincide...

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sorry for posting such a similar answer but after typing all this I couldn't just delete it. –  user50336 Nov 24 '12 at 17:23

Having a bit more exposure and experience now, I thought perhaps I should add this reflection as an answer for the sake of anyone who uses this question in the future.

Though it may be a bit unorthodox, I find the the fact that finite products and finite coproducts correspond in $\text{Ab}$ more obvious once on works through the fact that products and coproducts in $R\text{-mod}$, the category of $R$ modules, correspond. Once one has done this, one can use that fact that the categories $\text{Ab}$ and $\mathbb{Z}\text{-mod}$ are the same to conclude that finite products and finite coproducts in $\text{Ab}$ correspond.

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