# If $gi_j = fi_j$ for all coproduct inclusions $i_j$ of $\amalg_{j\in J}A_j$ then $f=g$

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.

• 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. – HallaSurvivor Oct 19 at 21:54
• @HallaSurvivor That should be an answer. Actually, you could just copy and paste this text. – Mark Kamsma Oct 19 at 22:35

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.