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.

Are all $\Pi A_\alpha \stackrel{\pi_i}\longrightarrow A_\alpha$ projection maps epic, given that $\Pi A_\alpha$ be the product of $A_\alpha$s? Of course, assuming that the product exists.

share|improve this question

4 Answers 4

up vote 4 down vote accepted

In an arbitrary category, it suffices to have a zero object $0$.

Consider $\pi_i\colon \prod A_j \rightarrow A_i$ and pick an index $n$. Then we have the identity morphism $\mathrm{id}_n\colon A_n\rightarrow A_n$ and to every other object $A_j$ we have the zero morphism $0_{nj}$. By the universal property of the product, we get a morphism $\rho_n\colon A_n\rightarrow \prod A_j$ with $\pi_n\circ \rho_n = \mathrm{id}_n$ and $\pi_j \circ \rho_n = 0_{nj}$ for all other $j$.

Now suppose there exists an object $X$ and $f,g\colon A_n\rightarrow X$ with $f\circ\pi_n = g\circ\pi_n$. Then

$$f\circ\pi_n\circ\rho_n = f\circ\mathrm{id}_n = f$$

and

$$g\circ\pi_n\circ\rho_n = g\circ\mathrm{id}_n = g$$

Since $f\circ\pi_n = g\circ\pi_n$, we get $f=g$.

The Wikipedia page on the product states that it is not true for arbitrary categories (without zero, therefore), but I'm not quick to find an example.

share|improve this answer
3  
You argument works even if the category does not have a zero object (for example in the category $\textbf{Set}$). It suffices that there exists at least one morphism between any two objects that make up the product. Then one can still construct a morphism $\rho_n$ which is a right inverse for $\pi_n$. And that implies that $\pi_n$ is an epimorphism. –  PatrickR Jan 26 '13 at 7:22

There are counterexamples even in Set: for any non-empty set $X$, the projection $\emptyset \times X \to X$ is not epimorphic. Assuming the axiom of choice, in Set, all counter-examples involve the empty set.

In a universe of sets where the axiom of choice fails, there are products $\prod_\alpha X_\alpha = \emptyset$ where none of the $X_\alpha$'s are empty sets; these would also give counterexamples.

share|improve this answer
    
Ah, of course, this is the most basic example. On the other hand, for sets (or topological spaces, or manifolds), always at least one projection is epic. For schemes this is not the case. –  Martin Brandenburg Jan 27 '13 at 14:18

By duality, the question is equivlent to: Are coproduct inclusions monic? The category of commutative rings provides many counterexamples, here $\sqcup = \otimes$ and $R \otimes 0 = 0$, so that $R \to R \otimes 0$ is not injective (unless $R=0$). A little bit more interesting, we have $\mathbb{Z}/2 \otimes \mathbb{Z}/3=0$, so that here both coproduct inclusions are not monic.

For the algebro-geometric minded reader: There are many non-empty schemes $X,Y$ such that $X \times Y = \emptyset$, so that the projections are not epic ... epic fail!

share|improve this answer

Judging by the tag I assume you are referring to the categorical product. In that case here is a small counter example showing that the categorical projections for a categorical product, when it exists, need not be epimorphic. Consider a category with objects $x,y,z$ and the following morphisms (other than the identities). There is precisely one morhpism $z\to x$ and one morhpism $h:z\to y$. It is immediate to verify that in that category $z$ is the product of $x$ and $y$. The projections here are epimorphic but that can easily be changed. Add now a fourth object $t$ together with two morphisms $f_{1,2}:y\to t$ and one morphism $g:z\to t$ with composition of these given by $f_i\circ h=g$. It is easily seen that $z$ is still the product of $x$ and $y$ (because nothing new has $z$ as codomain) but clearly the projection $h:z \to y$ is not an epimorphism. Of course you can tweak things some more to prevent the other projection from being epimorhpic.

share|improve this answer

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.