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Suppose I have some small coproduct $\amalg_{j\in J} A_j$ with coproduct inclusions $i_j:A_j\to \amalg_{j\in J}A_j.$ Suppose I have two maps $f,g:\amalg_{j\in J}A_j\to X$ for which $fi_j=gi_j$ for all $j\in J$. Does this imply that $f=g$? If we're working on sets it does, but I'm not sure how to show this when working in general categories. It seems very simple, but I have no clue where to begin.

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    $\begingroup$ Yes. Recall that the universal property of coproducts (and universal properties in general) requires there be a unique map out of the coproduct making the relevant diagram commute. Since $f$ and $g$ both make the diagram commute, they must be equal. $\endgroup$ – HallaSurvivor Oct 19 at 21:54
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    $\begingroup$ @HallaSurvivor That should be an answer. Actually, you could just copy and paste this text. $\endgroup$ – Mark Kamsma Oct 19 at 22:35
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Yes. Recall that the universal property of coproducts (and universal properties in general) requires there be a unique map out of the coproduct making the relevant diagram commute. Since $f$ and $g$ both make the diagram commute, they must be equal.


I hope this helps ^_^

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    $\begingroup$ But of course! How could I have missed it? Thank you so much! $\endgroup$ – Dasherman Oct 19 at 22:41
  • $\begingroup$ It happens to all of us ^_^ Happy to help! $\endgroup$ – HallaSurvivor Oct 19 at 22:44

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