The definition of product in Lang's Algebra (page 58) is this:

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Let $(P,f,g)$ be a product of $A$ and $B$. Let be $C=A$ and $\varphi=id_A$. Then, by definition, there is a (unique) $h:A\to P$ morphism so, that $id_A=f\circ h$, that is, $h$ is a right inverse of $f$, that is, $f$ is a retraction. The similar holds for $g$.

Is this proof correct? I'm pretty sure that yes, but I'm a bit surprised that I didn't find this statement anywhere. That's why I need a confirmation.

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    $\begingroup$ So your $\varphi$ is the identity. What's your $\psi$? $\endgroup$ Oct 7, 2018 at 7:36

2 Answers 2



It only works if some arrow $A\to B$ indeed exists.

If we are e.g. working in the category of sets with $B=\varnothing$ and $A\neq\varnothing$ then this is not the case and also $P=A\times B=\varnothing$.

In that case $f:P=\varnothing\to A\neq\varnothing$ has no right-inverse.


That seems ok, as long as there are arrows $A \to B$, because the existence of $h$ will be guaranteed provided that you give morphims $C \to A$ and $C \to B$, in which case $\phi = fh$ and $\psi = gh$ ought to exist.

If you have any arrow $a : A \to B$ then $(id_A,a)$ factors through $P$ via some unique $h$ as you have said.

As drhab has said in his answer, there are rather elementary examples in which this fails already. I know little category theory, but maybe more 'interesting' examples can be fabricated out of well known objects with no arrows between them.

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    $\begingroup$ The problem is that two fields of different characteristics cannot have a product (precisely because no field can have morphisms into the two field). $\endgroup$
    – Arnaud D.
    Oct 7, 2018 at 16:19
  • $\begingroup$ @ArnaudD. I should've seen that coming... edited accordingly, thanks. $\endgroup$
    – qualcuno
    Oct 7, 2018 at 18:14

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