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Exercise $6 (b)$, page 58 from Hungenford's book Algebra.

Show that in $\mathcal{S}_{\star}$ (the category of pointed sets) every family of objects has a coproduct (often called a "wedge product"); describe this coproduct.

I need a suggestion in order to find the coproduct. I would appreciate your help.

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With two normal sets, the coproduct is the disjoint union. With pointed sets, you merely add the condition that the basepoints of both sets always go to the basepoint of the new set, which only requires a small modification to the disjoint union. –  Carl Feb 23 '12 at 21:13
    
@Carl: I would like to thank you. Can you please write it as an answer so that I can accept it? Thank you again! –  spohreis Feb 23 '12 at 21:33
    
No problem! Reposting as an answer. –  Carl Feb 23 '12 at 22:03
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up vote 4 down vote accepted

HINT: With two normal sets, the coproduct is the disjoint union. With pointed sets, you merely add the condition that the basepoints of both sets always go to the basepoint of the new set, which only requires a small modification to the disjoint union.

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could you clarify what you mean by: "the basepoints of both sets always go to the basepoint of the new set"? Even better, could you please compute the coproduct of Sa={{a,b},a} (set {a,b} with base point a) and Sc={{c,d},d} (set {c,d} with base point c), if it exists? –  magma Feb 25 '12 at 14:29
    
@magma: It's just the disjoint union $X = \{a,b\} \coprod \{c,d\}$ with the base points $a,c$ identified, ie. the quotient of $X$ by the equivalence relation generated by $a \sim c$. –  Najib Idrissi Feb 25 '12 at 16:32
    
@magma: the coproduct of pointed sets $(X_i,b_i)$ is the universal pointed set $(X,b)$ with inclusion maps $f_i:(X_i,b_i)\to (X,b)$. By definition of morphisms between pointed sets, $f_i(b_i)=b$: all basepoints are mapped to the new basepoint. –  wildildildlife Feb 25 '12 at 17:15
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