Category-Theoretic Product

[ydoaPs]

I'm not sure I understand exactly the product in category theoretic terms. The book I'm using gives a very concrete example that is easy to understand. However, I'm not sure I get it generally. How does it work for the product of the reals?

 

 

What's X in that case? What are f₁ and f₂?

[ajb]

I would imagine X is any element of the category in question. You then have to think about how you are dealing with the real numbers in order to understand what the morphisms are.

 

Is that page of the book available to view on googlebooks or something similar?

[ydoaPs]

I would imagine X is any element of the category in question. You then have to think about how you are dealing with the real numbers in order to understand what the morphisms are.

 

Is that page of the book available to view on googlebooks or something similar?

Oh, yes, it is *any* object in the category. So, the point of having the arbitrary object and the 'extra' processes is to ensure that the product looks the same to every object in the category?

 

To answer my own question from the OP, if X were R (since we know R is an object in our category), then f₁=f₂=I and f=p₁^-1=p₂^-1.

Is that page of the book available to view on googlebooks or something similar?

That section is on neither the Amazon Look Inside nor the Google Books preview.

[Xerxes]

In diagrams of this sort, it is importat to "start" in the right place, in this cas it is [math]R^2[/math]

 

Assume we are working in the category of fields Fld.where [math]R[/math] dentes the Real field Then the projections (morphisma) [math]p_1 \to R[/math] and [math]p_2 \to R[/math] define the morphisms [math]R^2 \equiv R \times R \to R[/math].

 

And if there exist another pair of morphisms [math]f_1:X \to R[/math] and [math]f_2: X \to R[/math] then this induces the morphism [math]X \to R^2[/math].(Personally in this circumstance I orefer to use a dashed arrow for the induced morkisms - it's a matter of taste)

 

In Fld an example would be [math]X = C[/math], the complex field

[ydoaPs]

In diagrams of this sort, it is importat to "start" in the right place, in this cas it is [math]R^2[/math]

 

Assume we are working in the category of fields Fld.where [math]R[/math] dentes the Real field Then the projections (morphisma) [math]p_1 \to R[/math] and [math]p_2 \to R[/math] define the morphisms [math]R^2 \equiv R \times R \to R[/math].

 

And if there exist another pair of morphisms [math]f_1:X \to R[/math] and [math]f_2: X \to R[/math] then this induces the morphism [math]X \to R^2[/math].(Personally in this circumstance I orefer to use a dashed arrow for the induced morkisms - it's a matter of taste)

 

In Fld an example would be [math]X = C[/math], the complex field

In the complex field example, f₁=f₂ is a mapping from the real parts of the complex field to the corresponding parts of the real field, essentially chopping off the imaginary parts of the points. Is there any way to think about the induced morphism in this case other than just the composite p₂^-1(f₁)?

[Xerxes]

In the complex field example, f₁=f₂ is a mapping from the real parts of the complex field to the corresponding parts of the real field

Why do you think [math]f_1=f_2[/math] in this case?,.

Is there any way to think about the induced morphism in this case other than just the composite p₂^-1(f₁)?

Yes - you rather missed the point. We require your diagram to commute. That is the composite, say, [math]g_1:C \to R^2[/math] exists such that [math]p_1 \circ g_1 = f_1[/math]. Likewise for the "bottom half" of your diagram. Of course the morphism [math](g_1,g_2):C \to R^2[/math] makes no sense notationally, so call it anything you want.

 

In this case it is, of course an isomorphism, though I am not sure this is always the case

[ydoaPs]

Why do you think [math]f_1=f_2[/math] in this case?,.

In both cases, they're C->R, and that's a really simple process between them. If the product were RxZ, they wouldn't be the same process. Why would we want two different processes from C to R for the product?

[Xerxes]

Well I am not sure what you mean by a "process" - let's be conventional and call it a morphism. Assuming I was correct and we are working in the category of fields, then our morphisms are field homomorphisms.

 

Now I will say that 2 such morphisms are equal if and only if they have the same image and the same pre-image.

 

Now consider the "central" object in your diagram [math]R^2 \equiv R \times R[/math]. The projections [math]p_1((x,y)) = x \in R[/math] and [math]p_2((x,y)) = y \in R[/math] always exist.

 

And if [math]f_1=f_2[/math] then their images are equal, say to [math]x[/math]. This mandates that [math]p_1=p_2[/math] (else your diagram makes no sense in any category) in which case your diagram is completly trivial.

 

Otherwise we must have that [math]f_1 \ne f_2[/math]

 

Hmm....is this clear?