# Coproduct in ${\bf Set}$ of the same object

In a category $$\mathfrak{C}$$ (for instance $${\bf Set}$$) I want to compute the coproduct of the objects $$X$$ and $$X$$. Since they are not disjoint, I can consider these sets as two disjoint copies of the same sets and, next, compute their coproduct. Now, using the definition of coproduct, it is immediate to verify that $$(X,1)$$, where $$1$$ denotes the identity morphism. So, there must be an isomorphism between $$X$$ and the coproduct of $$X$$ with itself. Which is such an isomorphism?

• There needn’t be such an isomorphism. There are two maps $i_1,i_2: X\to X\sqcup X.$ There is also a map $p: X\sqcup X\to X,$ But these maps are not isomorphisms. – Thomas Andrews Mar 25 at 15:23
• But both of them are a coproduct of $X$ and $X$ – TheWanderer Mar 25 at 15:26
• No, $X$ is not the coproduct of $X$ with $X$. To verify this, let $A=\{1,2\}$, and define the two maps $f_1\colon X\to A$ sending everything to $1$, and $f_2\colon X\to A$ sending everything to $2$. If $X$ were the coproduct of $X$ with itself, then these two maps should induce a unique map $F\colon X\amalg X\to A$ such that $f_1=F\circ i_1$ and $f_2=F\circ i_2$. You are claiming $i_1=i_2=\mathrm{id}_X$ and $X\amalg X=X$. So you are claiming that $f_1=F=f_2$. That is just not true. – Arturo Magidin Mar 25 at 15:35
• Go back tot he axiom of coproduct and argue how $X$ is the coproduct. It doesn’t work. – Thomas Andrews Mar 25 at 15:41

No, it's not true unless $$X$$ is the initial object.
The coproduct of a singleton set $$X$$ with itself will contain $$2$$ elements, which is not isomorphic to $$X$$
• No, $X$ does not satisfy. – Berci Mar 25 at 15:36