Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Right now I am reading "Topoi: The Categorial Analysis of Logic", by Goldblatt. I am at the section where he explains the concept of products, by first looking at an example from Set (not using specific sets, just an abstract A and B) and then defining what a product in a category is.

So in getting the concept down I figured I would do actual examples and even non-examples myself. His second example is that in Grp, the category of groups, a product is the direct group product of groups, the binary operation done component-wise. He doesn't use specific groups, just says that the direct group product is a categorial product.

Well I figured that $Z_8$, not being isomorphic to $Z_2$ X $Z_4$ in Grp (though as merely sets are bijective), would be a great non-example of a product (since products are all merely isomorphic anyway they form a class of objects, right? so there might be more than one of them?) I had trouble doing commutative diagrams in latex but this was my set up:

$pr_2\colon Z_2$ X $Z_4 \to Z_2$

$pr_4\colon Z_2$ X $Z_4 \to Z_4$

And here, my "auxilary" object and its homomorphisms:

$f\colon Z_8 \to Z_2$

$g\colon Z_8 \to Z_4$

$(f,g)\colon Z_8 \to Z_2$ X $Z_4$

Where f could either be the trivial homomorphism or x mod 2, and similarly for g.

But no matter what I fill in for f and g, (f,g) seems to not only be defined and commute the diagram, but unique. However, given the group homomorphisms from $[Z_8, Z_2]$ and $[Z_8, Z_4]$, (f,g) is clearly not an isomorphism even if we were to forget the group structure, but not iso even as a homomorphism, yet the product arrow is supposed to be iso.

When first coming to the concept from the book days ago, I was confused if whether or not the auxilary object too was a product but I was confident that I understood that it too is a product, and confident that I was following the logic of the author's text in the language of category theory just by following the "mechanical manipulation of symbols", for example on the next page he proves that all product objects are iso to each other and I felt I was able to follow his argument.

I guess my question is, what am I not understanding in the definition of a product in category theory? How does it entail the product from specific categories such as Grp and Top, and further properties like the product arrows being iso's in the specific categories? And how would that explanation/understanding correct whatever mistake I made with the above example?

Thanks in advance.

share|improve this question

closed as off topic by Andreas Blass, Dan Petersen, Tom Leinster, Steven Landsburg, Daniel Moskovich Mar 12 '12 at 2:18

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

$Z_8$, which I believe is what you refer to as "the auxiliary object," is not a product of $Z_2$ and $Z_4$, no matter what $f$ and $g$ you use. –  Andreas Blass Mar 11 '12 at 22:50
It is questionable that this is the right site for your question; MO is mainly for questions of interest to research mathematicians. But in brief, you seem to have gotten the idea that (f, g) must be an isomorphism. It doesn't. It's only an isomorphism if the data of f and g are the projections of a product structure. Anyway, you might try asking your question at math.stackexchange.com. –  Todd Trimble Mar 11 '12 at 22:53
Thanks for the alternate website, and I thought I was clear about $Z_8$ being a non-example of product i.e. it NOT being a product. I stuck it in the diagram on purpose, knowing in advance it isn't iso to $Z_2$ X $Z_4$, believing that studying the diagram would not only show it not be iso but also believing I wouldn't be able to make a commutative diagram. But as I said, it does seem to fit, which countered my expectations leaving me to doubt my understanding of a product in the language of category theory. I wanted to simply know what I was missing. –  Chris Mar 11 '12 at 23:02
Chris, I'm voting to close. You got an answer that seems to have helped, but really your question isn't appropriate for this site. That doesn't mean it's a bad question, just that it doesn't belong here. –  Tom Leinster Mar 11 '12 at 23:42

1 Answer 1

up vote 2 down vote accepted

When first coming to the concept from the book days ago, I was confused if whether or not the auxilary object too was a product but I was confident that I understood that it too is a product,

If I've understood this sentence, this is what you are not understanding.

The definition of a product is that for any object $C$ and arrows $f:C\to A$ and $g:C\to B$, there is a unique arrow $(f,g):C\to A\times B$ making the diagram commute. In particular, the object $C$ has no reason to be a product. When he proves that products are unique up to isomorphism, he just applies the universal property taking for $C$ another product of $A$ and $B$, but in general you can apply it for any object $C$ (and to prove that something is a product, you have to verify it for every $C$)

In your example, the fact that you find that there is a unique arrow $(f,g)$ is reassuring, this is just a consequence of the fact that $Z_2\times{}Z_4$ is indeed a product of $Z_2$ and $Z_4$. If you want to prove that $Z_8$ is not such a product, you have to prove that for every pair of maps $pr_1:Z_8\to Z_2$ and $pr_2:Z_8\to Z_4$, there exists a group $C$ and maps $f:C\to Z_2$ and $g:C\to Z_4$ such that there is no arrow (or more than one) $C\to Z_8$ making the diagram commute.

share|improve this answer
Ah thanks. That makes much more sense. Ok so the iso products form a class of objects that not only fit the diagram at top, the 'auxilary' object, but for it to be a product it must also fit the 'middle base object'. There exists a (perhaps larger) class of objects that may fit the auxilary object for the product in the middle base, but they may not themselves work in the base. So while Z8 will commute the diagram if at top, it will not at bottom. But all isomorphic products will fit both the top and bottom. Right? –  Chris Mar 11 '12 at 23:33

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