         # Productmorphism

### Definition

Let $$f:X\rightarrow A$$ and $$g:Y\rightarrow B$$ then the productmorphism $$f\times g: X\times Y\rightarrow A\times B$$ is defined with $$q_1:X\times Y\rightarrow X$$, $$q_2: X\times Y\rightarrow Y$$, $$p_1:A\times B\rightarrow A$$ and $$p_2:A\times B\rightarrow B$$ as well as the relationships $$f\circ q_1= p_1\circ (f\times g)$$ and $$g\circ q_2= p_2\circ (f\times g)$$ as follows: $$f\times g= \langle f\circ q_1, g\circ q_2\rangle$$.

This relationship becomes clear when looking at the commutative diagram for multimorphisms If we replace $$Z$$ by $$X\times Y$$ and integrate $$q_1:X\times Y\rightarrow X$$ and $$q_2:X\times Y\rightarrow Y$$ as intermediate steps over $$X$$ and $$Y$$ respectively, we obtain the following diagram: ## Co-Productmorphism

As with multimorphism, the dual concept of co-productmorphism also exists for the productmorphism

We obtain the co-productmorphism analogous to the above by replacing $$Z$$ with $$X+Y$$ in the commutative diagram of the polymorphism and then introducing $$p_1=X\rightarrow X+Y$$ and $$p_2=Y\rightarrow X+Y$$ as intermediate steps via $$X$$ and $$Y$$, respectively. As a result we get the following diagram ### Definition

Let $$f:X\rightarrow A$$ and $$g:Y\rightarrow B$$ then the co-productmorphism $$f+g:A+B\rightarrow X+Y$$ is defined with $$q_1:A\rightarrow A+B$$, $$q_2: B\rightarrow A+B$$, $$p_1:X\rightarrow X+Y$$ and $$p_2:Y\rightarrow X+Y$$ as well as the relationships $$p_1\circ f= (f+g)\circ q_1$$ and $$p_2\circ g=(f+g)\circ q_2$$ as follows: $$f+g=[p_1\circ f, p_2\circ g]$$.

The only difference to the polymorphism is that the co-productmorphism maps to the same co-domain in both cases.

## Preview

Next time you will get everything about F-algebras, so we will have all the basics together to define recursive morphisms... exciting! 