Are the canonical injections into the coproduct a structure or a property?

Question

Can someone illuminate why the choice of canonical injections seems to be a part of the structure of a coproduct? Given a category $\mathcal{C}$ and objects $A, B$. Then their coproduct, if it exists, is defined to be the triple $(A\coprod B,\iota_A\colon A\to A\coprod B, \iota_B\colon B\to A\coprod B)$ subject to the universal property that given $f_A\colon A\to C, f_B\colon B\to C$ there exists a unique $f\colon A\coprod B\to C$ such that $f_A=f\circ\iota_A, f_B=f\circ\iota_B$.

I wonder if it suffices to demand that there are just some maps, say, $j_A\colon A\to A\coprod B, j_B\colon B\to A\coprod B$ such that given $f_A\colon A\to C, f_B\colon B\to C$ there exists a unique $f\colon A\coprod B\to C$ such that $f_A=f\circ j_A, f_B=f\circ j_B$.

For example in the category of groups: Is the free product of two groups $G,H$ just the group $G\ast H$ for which there exist some maps (the canonical inclusion maps $\iota_G, \iota_H$) that satisfy the universal property or should the coproduct be the triple $(G\ast H, \iota_G, \iota_H)$? I was taught the latter but I am not sure about the significance.

Maybe it has some significance for being unique up to unique isomorphism but I am puzzled here and would appreciate your help.

Answer 1

If you only demand that there are some such maps, you cannot distinguuish $A\oplus B$ from $B\oplus A$, for instance. In the end, the canonical maps used in the universal peroperty are the only thing you have and know about the product, coproduct, limit, colimit, hence it is important to carry them with you ...

Answer 2

The problem is that the construction of $f$ from $f_A,f_B$ depends on $i_A$ and $i_B$. So if you construct different maps from a coproduct using the universal property, you need to make sure that you always use it on the same pair of injections.

If I give you, for example, a map $g:C\to D$, you would have maps $h_A=g\circ f_A:A\to D$ and $h_B=g\circ f_B:B\to D$, and you could construct a map $h:A\sqcup B\to C$ from there. Then we can show that $h=g\circ f$... but only if $f$ and $h$ were constructed using the same $i_A,i_B$ !

A similar example : if we have maps $\alpha:A'\to A$ and $\beta:B'\to B$, then we can define a map $\alpha\sqcup \beta : A'\sqcup B'\to A\sqcup B$. If we also have maps $\alpha':A''\to A'$ and $\beta':B''\to B'$, then $$(\alpha\sqcup \beta)\circ (\alpha'\sqcup \beta')=(\alpha\circ \alpha')\sqcup (\beta\circ \beta'),$$ and moreover $$1_A\sqcup 1_B=1_{A\sqcup B}.$$ So the coproduct is functorial... but again, all this only holds if we always use the same maps $i_A,i_B$ to define $A\sqcup B$ (and similarly for $A'\sqcup B'$ and $A''\sqcup B''$).

Answer 3

Yes, you're exactly right that a coproduct object for $A$ and $B$ could be defined as an object $C$ for which there exists a pair of injections with the universal property.

This definition would be equivalent to the existing definition. It would, however, miss out on the fact that in category theory, arrows are the focal point. This is a key strength of category theory, and so we do want to place focus on arrows.

For example, a focus on arrows exposes the "forgetful functors" that we always use but seldom recognize; it exposes the maps that characterize co/equalizers; it gives us element-free characterizations of injective and surjective functions, and function-based ways to count the elements or subsets of a set; it tells us what's special about $\varnothing$ and singleton sets. Though sometimes we mention objects anyways, it is the arrows that are most relevant.

In this light, what is important about the coproduct is not the object for which some arrows exist; it is the arrows themselves. So, if we are to leave anything out of the definition of coproduct, I would de-emphasize the objects:

A coproduct is a pair of injections $i,j$ into the same codomain which is moreover universal with respect to that property: if $i_2, j_2$ are any arrows that share a codomain and have the same respective domains as $i$ and $j$, then there exists a unique arrow $!$ such that $!\circ i = i_2$ and $!\circ j = j_2$.

The injections $\langle i,j\rangle$ carry all the data you need to completely define "the" coproduct.