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I wonder if given an injection $A^2 \rightarrow B^2$, there necesarrily exists an injection $A \rightarrow B$. And if there is, can you construct it canonically? If we assume the axiom of choice, then non-existence of an injection implies existence of a surjection. So then this can be rephrased as given a surjection $A \rightarrow B$, construct a surjection $A^2 \rightarrow B^2$, which off course is possible canononically. So my question is if we can find a canonical injection $A \rightarrow B$ given an injection $A^2 \rightarrow B^2$ without invoking the axiom of choice.

Jelle Bloemendaal
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    I don't see why would there be a canonical choice of injection. Let $A = {0}$ and $B = {0,1}$. Then $f(0,0) = (0,1)$ is injection $A^2\to B^2$. How would you define $\bar f(0)$ then? As $0$ or $1$? – Ennar Nov 21 '23 at 11:11
  • It seems that this should be true from a simple cardinal argument: if $A^2$ has fewer elements than $B^2$ then certainly $A$ has fewer elements than $B$. But constructing the injection from $f$ without the axiom of choice doesn't look particularly easy. If $f = (f_1, f_2)$, then the injectivity of $f$ certainly does not imply any injectivity in $f_1$ nor in $f_2$. – Stef Nov 21 '23 at 11:28
  • @Ennar, may be canonical isn't the right word indeed. I mean a general recipe, just like how you can construct a bijection given an injection from $A$ to $B$ and $B$ to $A$. – Jelle Bloemendaal Nov 21 '23 at 13:36
  • @JelleBloemendaal you always have diagonal injection $A\to A^2$ that you can use but for $B^2\to B$ you need to pick projection and you can't do that consistently I think. – Ennar Nov 21 '23 at 14:46

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