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I am trying to find whether the polynomial (monomial) functor $P : X \rightarrow X\times X $, i.e. $P(X) = X^2$, is monomorphic on objects, in other words, that if there exists an isomorphism $A\times A \overset {i} {\hookrightarrow} B \times B$, then there is also an isomorphism $A \overset {j} {\hookrightarrow} B$.

I understand that it is important to use the fact that those are squares, i.e. products of isomorphic objects, else it doesn't hold (indeed, $A \times A = (A\times A)\times 1$, etc.).

However I have trouble to figure out the path of the proof. I'm not even sure the property holds outside of the category $Sets$.

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    $\begingroup$ That's not what 'faithful' means in my book. What you wrote looks closer to (but is not the same as) the notion of conservative functor: ncatlab.org/nlab/show/conservative+functor $\endgroup$ – Todd Trimble Oct 16 '17 at 20:50
  • $\begingroup$ Yes, thank you. Let me fix it. Let's say "mono on objects" for now. $\endgroup$ – Almeo Maus Oct 16 '17 at 21:00
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    $\begingroup$ Of course it holds in the category of sets. $\endgroup$ – HeinrichD Oct 30 '17 at 13:29
  • $\begingroup$ @HeinrichD Yes, and in Sets this property requires the axiom of choice (Tarski theorem) to hold. $\endgroup$ – Almeo Maus Nov 1 '17 at 23:14
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There is a group $A$ that is isomorphic to $A \times \mathbb Z \times \mathbb Z$ but not isomorphic to $A \times \mathbb Z$.

Taking $B = \mathbb A \times \mathbb Z$ produces a counterexample in the category of abelian groups.

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    $\begingroup$ see also the Boolean algebra counterexample mathoverflow.net/a/227525/18060 $\endgroup$ – Will Sawin Oct 16 '17 at 20:49
  • $\begingroup$ thank you so much!! This counter-example with groups is perfect. $\endgroup$ – Almeo Maus Oct 16 '17 at 20:56
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    $\begingroup$ Just a historical comment: Whether there are non-isomorphic abelian groups with $A\times A\cong B\times B$ was Kaplansky’s second “test problem” from 1954, and the first example was published by Jonsson in 1957, long before anybody looked for examples of abelian groups with $A\times\mathbb{Z}\not\cong A\cong A\times\mathbb{Z}\times\mathbb{Z}$. $\endgroup$ – Jeremy Rickard Oct 16 '17 at 22:23

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