In a category the product $X$ of a family $(X_i,x_i:X\rightarrow X_i)_i$ was of the form: For all $M$ and morphisms $f_i:M\rightarrow X_i$ there is a unique morphism $f:M\rightarrow X$ such that $x_i\circ f=f_i$ for all $i$.
Now I had the (maybe senseless) idea of changing the directions of the arrows in the corresponding diagrams.
What if I change the direction of arrows lets say in my favourite category Ab of abelian groups. So Is there something of the following:
- Given a family $(X_i,x_i:X_i\rightarrow X)_i$ in Ab. For every abelian group $M$ and grouphomomorphisms $(f_i:M\rightarrow X_i)_i$ there is a unique $f:M\rightarrow X$ such that $f=x_i\circ f_i$ for all $i$.
- Given a family $(X_i,x_i:X_i\rightarrow X)_i$ in Ab. For every abelian group $M$ and grouphomomorphisms $(f_i:X_i\rightarrow M)_i$ there is a unique $f:M\rightarrow X$ such that $x_i=f\circ f_i$ for all $i$.
- What if we look at $(X_i,x_i:X\rightarrow X_i)_i$ with $(f_i:X_i\rightarrow M)_i$?
As I have said, that could be a senseless question. In every case I am interested in the answer.
Thank you!
These things essentially never occur. In 1, suppose you're taking two groups $X_1,X_2$. We have the identity map $\mathrm{id}:X_1\to X_1$ and the zero map $0:X_1\to X_2$. We necessarily have $0=x_2\circ 0$, where the zero on the left is $0:X_1\to X$, so by uniqueness we must have $x_1\circ \mathrm{id}=0,$ that is $x_1=0$, and similarly for $x_2$. In fact, that the $x_i$ be zero is necessary and sufficient for 1; the uniqueness clause is vacuous since we explicitly require $f=x_1\circ f_1$. In particular, $X$ is very much not specified up to isomorphism! The problem is that you require $x_i\circ f_i=x_j\circ f_j$ for all $i,j$ and all maps $f_i$; this is an implausible thing to ask for, and the only way to get it is by making the $x_i$s degenerate.
The $x_i$ are also zero in 2; consider $M=0$. Similarly for 3, which is dual to 1.