Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can two morphisms (say of the category Set, but also in every category with binary product) be restored knowing a product of these two morphisms?

I'm especially interested in the case if one of the morphisms is empty, e.g. is the morphism from an empty set to an empty set.

share|improve this question
    
Your question is unclear. Do you want to understand morphisms $Y \to X_1 \times X_2$, morphisms $X_1 \times X_2 \to Y$, or morphisms $Y_1 \times Y_2 \to X_1 \times X_2$? –  Zhen Lin Mar 17 '12 at 14:33

1 Answer 1

By the very definition of a categorical product, $f_i = \pi_i \circ f$ where $f$ is the product of $f_1, f_2$, so it's always possible to recover the morphisms. If one of them is empty, for example $f_1 : Y \rightarrow X_1 \times X_2$ is empty, then it means that $Y = \varnothing$, and the other function must be the empty function too, and $f_1 \times f_2 = \varnothing$ too.

In general, if $Y$ is the initial object of $\mathcal{C}$ (like $\varnothing$ is the initial object of $\mathcal{Set}$), then in the commutative diagram defining the product, the morphisms $f_i : Y \rightarrow X_i$ are the only morphisms from $Y$ to $X_i$, and their product $f$ is the only morphism from $Y$ to the product of the $X_i$.

share|improve this answer
    
Sorry for a stupid question, but why $\pi_i$ are unique? –  porton Mar 17 '12 at 14:27
1  
They are not "unique", but they are part of the product: in category theory, the product of $X_1$ and $X_2$ is a object $X$ together with two morphisms $\pi_i : X \rightarrow X_i$ that make the usual commutative diagrams commute. See here for reference: en.wikipedia.org/wiki/Product_(category_theory) –  Najib Idrissi Mar 17 '12 at 14:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.