# Is it true, that every morphism in a product is a retraction?

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

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.

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

Almost.

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.

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