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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?

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  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Mar 25 at 15:23
  • $\begingroup$ But both of them are a coproduct of $X$ and $X$ $\endgroup$ – TheWanderer Mar 25 at 15:26
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Mar 25 at 15:35
  • $\begingroup$ Go back tot he axiom of coproduct and argue how $X$ is the coproduct. It doesn’t work. $\endgroup$ – Thomas Andrews Mar 25 at 15:41
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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 $X$

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  • $\begingroup$ It is obvious that these objects are not isomorphic, however, both of them satisfy the definition of coproduct. $\endgroup$ – TheWanderer Mar 25 at 15:28
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    $\begingroup$ No, $X$ does not satisfy. $\endgroup$ – Berci Mar 25 at 15:36

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