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Let $\mathfrak{C}$ be the class of all algebras of type $(0,0)$. Choose two nonisomorphic algebras $A$ and $B$ with two elements in $\mathfrak{C}$ and describe the coproduct of $A$ and $B$ in $\mathfrak{C}$.

Let me outline my thoughts and why I'm stuck: First of all the only two noisomorphic algebras that come to my mind are $A=(\{a_1,a_2\}, a_1, a_2)$ and $B=(\{b_1,b_2\}, b_1,b_1)$. Now assume I have a coproduct $\coprod$ with operations $(c_1, c_2)$ then $\iota_B$ has to map the two operations $b_1\mapsto c_1$, $b_1\mapsto c_2$, so there has to be $c_1=c_2$ otherwise I would have two images for element $b_1$. But with $c_1=c_2$ I see no options to make $f_A=f\circ \iota_A$ for arbitrary homomorphism $f_A: A\to C$ for arbitrary $C$ in $\mathfrak{C}$, as both $a_1$ and $a_2$ would be mapped to the same element in $C$. So I guess I am doing something fundamentally wrong here.

Thanks in advance for helping me resolve that thinking barrier

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I think there is no contradiction! Maps out of the coproduct correspond to pairs of maps out of $A$ and $B$. Of course, if there is a map $B \to C$, then $C$ must have both constants refer to the same element. But then maps $A \to C$ must also send the (distinct) constants in $A$ to the same point of $C$.

So it's completely ok for the constants to be identified in the coproduct, because anything which $A$ and $B$ can both map into will have the constants identified - indeed the coproduct must identify the constants to reflect this.


I hope this helps ^_^

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    $\begingroup$ Ah right, I totally missed that. So you agree the coproduct hast the form $(\{o,p\}), o,o)$ with with $\iota_A: a_1,a_2\mapsto o$ and $\iota_B: b_1 \mapsto o,\ b_2 \mapsto p$ or am I missing something again? $\endgroup$
    – GEO
    Commented Oct 24, 2019 at 7:34
  • $\begingroup$ HallaSurvivor: Could you leave feedback to my question before I mark your answer as accepted? $\endgroup$
    – GEO
    Commented Oct 24, 2019 at 17:50
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    $\begingroup$ Sorry, I upvoted your question as a way of saying yes. - I agree with you ^_^ $\endgroup$ Commented Oct 24, 2019 at 18:01

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