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Conceptually, I get the difference between products and coproducts: the first has projections, the second has inclusions. There are all sorts of circumstances in which you can be convinced that these two notions are different. But now I'm thinking about groups, and it seems to me that (direct) products of groups actually do come with inclusions as per the universal property:

The product $\prod G_\alpha$ of groups has projections $\{\pi_{\alpha_0}:\prod G_\alpha \to G_\alpha\}$ to each of its factors. Then for each $\alpha_0$, every collection of maps $\{f_\alpha:G_{\alpha_0} \to G_\alpha\}$, by $f_{\alpha_0} = \mathrm{id}$ and $f_\alpha \equiv e$ else, factors through a unique map

$$ G_{\alpha_0} \xrightarrow{\exists!\ i_{\alpha_0}} \prod G_\alpha \xrightarrow{\pi_{\alpha_0}} G_{\alpha_0}.$$

Clearly the map $i_{\alpha_0}$ is an embedding, so I'd like to think of it as an inclusion. And this is all very natural, basically because groups are special in that they all have a distinguished element (the identity). So is there some nice conceptual reason why the collection $\{i_{\alpha_0}:G_{\alpha_0} \to \prod G_\alpha\}$ does not make $\prod G_\alpha$ into the coproduct, aside from the fact that of course the direct product and free product of groups are not isomorphic?

Remark: I suppose you can also do a similar sort of thing to show that the coproduct also has projections, by basically dualizing this argument.

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    $\begingroup$ Actually, in the category of Abelian groups, for a finite index set, it does coincide with the coproduct. $\endgroup$ – Berci Mar 28 at 19:03
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    $\begingroup$ @Berci right, but what on earth makes it fail in Grp? I mean sure, the Cartesian product has some sort of commutativity relation built in, which contradicts the „no unnecessary relations“ mantra of free groups (left adjoints preserve coprods). But this doesn’t really explain, why the Cartesian product is not a coproduct in the presence of somewhat canonical choices of inclusions (like for monoids, groups or pointed sets)... $\endgroup$ – PrudiiArca Mar 28 at 19:13
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    $\begingroup$ Well, I think, what you say in some sense does explain why it's not the coproduct. The key is that in the presence of a free-forgetful adjunction, the coproduct must be the corresponding free construction, namely if the objects are given by their (group) presentation, then the disjoint union of these is a presentation of their coproduct. In particular, if $F(X)$ is the free object over set $X$, then $F(X)\sqcup F(Y) =F(X\sqcup Y)$, because $F$ preserves colimits. $\endgroup$ – Berci Mar 28 at 19:19
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    $\begingroup$ What you discuss is only one part of the universal property, the other part is the relation to other objects in the category. Morphisms from the product induce morphisms on the factors, but morphisms on the factors does not necessarily induce one on the product. Is that conceptually what you are looking for? Also hi. $\endgroup$ – Paul Plummer Mar 28 at 19:27
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    $\begingroup$ Part of the difficulty might lie in the first line of the question, where "the difference" should be "a difference", because it's not the only difference. The universal properties also matter. In particular, the universal property of a coproduct implies that the inclusions are jointly epic. That's not the case for products of groups (if infinitely many factors are nontrivial). $\endgroup$ – Andreas Blass Mar 29 at 2:45
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I feel like the key observation here is that in some categories (like groups) we have a canonical map from coproducts to products, induced by these inclusion maps that you've noticed exist. The natural question is where does it come from? Perhaps answering this question will shed some light on what's going on here, since we can then take a look at other examples of this phenomenon to get intuition.

Motivation: (This section is sorta fuzzy mathematically in places to motivate the definitions without worrying about details)

So what is a map from the coproduct to the product? Well by definition, if $\newcommand\C{\mathcal{C}}\C$ is our category, then $$ \newcommand\of[1]{\left({#1}\right)} \C\of{ \coprod_i X_i, \prod_j X_j } \cong \prod_{i,j}\C(X_i,X_j). $$ So a map from the coproduct to the product requires for every pair of objects a choice of map $\C(X,Y)$. When $X=Y$, this is easy, we can take the identity map. What do we do if $X\ne Y$ though? Well, if we require our category has finite products/coproducts including initial and terminal objects, $0$ and $1$ respectively, then if the unique map $0\to 1$ is an isomorphism, then we can always produce a map $$X\to 1\to 0\to Y,$$ and this doesn't depend on our choice of $0$ or $1$, since everything is unique up to isomorphism.

This gives us a definition.

Categories with zero objects:

If $0\to 1$ is an isomorphism, then $0$ is both initial and terminal, and we say $\C$ has a zero object, and from now on I'll write $0$ for a zero object.

We also say that the unique map $X\to 0\to Y$ is the zero morphism from $X$ to $Y$, written as $0$. (This gives a canonical enriching in pointed sets with smash product, which is related to your observation that groups have a distinguished element).

Thus in a category with zero objects, we can define a canonical morphism $$ \coprod_i X_i \to \prod_j X_j, $$ with components $1_{X_i}$ when $i=j$ and $0$ when $i\ne j$. (Writing this as a matrix, you'll note that this is the identity matrix).

However, this canonical map is not generally an isomorphism. When it is (for finite sums/products), we call the object the biproduct, written $X\oplus Y$, and in such a situation, we get a canonical enriching in commutative monoids. The addition of $f,g : X\to Y$ is given by the composite $$ X\xrightarrow{\Delta} X\oplus X \newcommand\toby\xrightarrow\toby{f\oplus g} Y\oplus Y \toby{\nabla} Y. $$

Examples:

The trivial group, $1$, is the zero object in both groups and abelian groups, and you can check that the canonical morphism I define above gives you the same result as the morphism induced by the universal property given by your inclusions. I.e., it sends $g \in G_i$ to the tuple $(1,1,\ldots,1,g,1,\ldots,1)$, with $g$ in the $i$th spot.

For another category, you have the category of pointed sets or pointed topological spaces (pairs $(X,x)$ with $x\in X$ and morphisms $f:(X,x)\to (Y,y)$ are the maps $f:X\to Y$ such that $f(x)=y$).

The coproduct here is called the wedge sum, and it naturally includes into the product in the same way, with $x\in X_i$ mapping to $(*,\ldots,*,x,*,\ldots,*)$, where $x$ is in the $i$th location, and $*$ is the basepoint in the other factors.

Finally, let's give a bit of a weird one. (Though this is a category I came across recently.)

$R$-algebras (for me right now, the category of commutative, unital algebras over a commutative ring $R$) do not have zero objects (unless $R=0$). The initial object is $R$, and the terminal object is the zero ring. However, we can consider the category of $R$-algebras with augmentations. Explicitly, these are commutative rings $S$ with maps $$ R\toby{\iota_S} S \toby{\pi_S} R,$$ where $\pi_S\iota_S = \newcommand\id{\mathrm{id}}\id_R$. Morphisms are ring maps $\phi : S\to T$ such that $\pi_T\phi = \pi_S$ and $\phi\iota_S = \iota_T$. Now $R$ is a zero object in this category. (This is a special case of a general way to produce new categories with zero objects, make the objects either pairs $(X,1\to X)$ of an object and a morphism from the terminal object or the dual construction, take pairs $(X,X\to 0)$, which is what we did here). The first construction is common, and called taking the pointed category of $\C$, denoted $\C_*$, and is what we do to produce pointed sets and pointed topological spaces.

The coproduct of $S$ and $T$ is $S\otimes_R T$, with augmentation given by $(s\otimes t)\mapsto \pi_S(s)\pi_T(t)$. The product of $S$ and $T$ is given by (the fiber product) $S\times_R T$, with algebra structure map given by $r\mapsto (\iota_S(r),\iota_T(r))$.

Then the morphism $S\otimes_R T\to S\times_R T$ is given by $s\mapsto (s,\iota_T(\pi_S(s))$ and $t\mapsto (\iota_S(\pi_T(t)),t)$.

Conclusion

Hopefully I've given a bit of context and background to the construction that you've noticed. I hope you'll see that asking the question of why the product fails to be the coproduct in groups, while a reasonable question, doesn't have much of an answer other than: because it can't be.

While that can feel very unsatisfying as an answer, I hope that seeing that there are lots of similar examples will make the following analogy make sense.

The inclusions produce a map from the coproduct to the product, in my analogy, I want to think that this is like proving an inequality. And sometimes this inequality is a strict inequality (the map is not an isomorphism), but sometimes in special cases, the inequality is an equality (the map is an isomorphism), and then something special and interesting happens. But the point is that asking why something is a strict inequality is hard to answer with anything other than because we can prove that they aren't the same. Instead, asking when do we have equality may be more fruitful. (Not a perfect analogy, granted.)

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    $\begingroup$ In this old answer, I characterized those varieties of algebras such that the factors of a product embed in the product. I'm putting the link here, because it seems closely related to your answer. $\endgroup$ – Alex Kruckman Mar 29 at 16:02
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Following your remarks about commutativity in the comments, let's cook up an example where the extra commutativity relations in $A\times B$ are the reason why it can't be a coproduct of $A$ and $B$.

For this, I'm going to want a pair of elements of some group that don't commute. Let's take $S_3$ the group of symmetries of a three-element set, and $c$ a 3-cycle in that group and $t$ a transposition (it doesn't matter which ones). We have $ct \not= tc$.

Now let's just take the coproduct of $S_3$ with itself, $S_3\sqcup S_3$. Let's call the inclusions $l$ and $r$. From the universal property of $S_3 \sqcup S_3$, whenever there are morphisms $f,g:S_3\to R$, there is $h: S_3\sqcup S_3 \to R$ with $hl = f$ and $hr = g$.

Specifically, let's pick $R = S_3$ and $f = g = \mathrm{id}$. So there should be $h : S_3\sqcup S_3 \to S_3$ such that $hl = hr = \mathrm{id}$. Now, that means that $hl(c)$ and $hr(t)$ don't commute, since they're just $c$ and $t$. But that means that $l(c)$ and $r(t)$ can't commute either, since applying a homomorphism to a commuting pair gives you a commuting pair. But with the coproduct as you describe with the inclusions you describe, they do commute, meaning that we can't even define $h$, let alone check it for uniqueness.

I'm not exactly sure how to communicate my intuition on this, but it's something like: whenever you can map a collection of $A_i$ in some target space $R$, do some operations with them there, and look at the result, then you can interpose the coproduct between the $A_i$ and the $R$, do the operations there instead, and get the same result. In particular, this means that anything that you can determine about which $A_i$ were inserted and which operations were done by looking at the result in $R$ must also be "remembered" by the coproduct. The direct product forgets the order in which you multiplied the elements of different $A_i$, so it won't always be possible to interpose it when that order is relevant to the result.

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