# Coproduct in class of algebras with two nullary operations

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

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 ^_^

• 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?
– GEO
Commented Oct 24, 2019 at 7:34
• HallaSurvivor: Could you leave feedback to my question before I mark your answer as accepted?
– GEO
Commented Oct 24, 2019 at 17:50
• Sorry, I upvoted your question as a way of saying yes. - I agree with you ^_^ Commented Oct 24, 2019 at 18:01