4
$\begingroup$

Suppose you have two morphisms $f\colon A\to B$ and $g\colon A\to C$. Then the fiber coproduct of $B$ and $C$ over $A$ exists and is the disjoint union $B\coprod C$ where we identify $fx$ and $gx$ for all $x\in A$.

I'm a little confused on how this satisfies the universal property. Suppose you had another set $D$ and morphisms $h\colon B\to D$ and $k\colon D$ such that $hf=kg$, so that the diagram commutes. How would you define the unique morphism $u$ from $B\coprod C\to D$?

Is it something along the lines of $u(b)=h(b)$ if $b\notin f(A)$ and $u(c)=k(c)$ if $c\notin g(A)$, and $u(f(x))=u(g(x))=k(f(x))=h(g(x))$ if $f(x)=g(x)$ is one of the points we smashed together in $B\coprod C$? And this definition makes sense since $hf=kg$.

$\endgroup$
3
$\begingroup$

That is exactly it. Can you see why the map is unique?

(Perhaps it is better to call this just the "pushout" of $B \leftarrow A \rightarrow C$, or more informally the "glueing of $B$ and $C$ along $A$". The name "fibre coproduct" seems a bit weird, because the construction has nothing to do with fibres!)

$\endgroup$
4
  • 1
    $\begingroup$ Although, the pullback is often times called the "fibered product" where there, for example in schemes, it does have to do with fibers. Then, by the convention of "co"ing things, there is a basis to call it the fibered coproduct :) $\endgroup$ – Alex Youcis Nov 16 '13 at 7:34
  • 2
    $\begingroup$ It should be a cofibred coproduct, then! $\endgroup$ – Zhen Lin Nov 16 '13 at 9:01
  • $\begingroup$ @AlexYoucis I undestand why someone would call it the "fibred coproduct", but I disagree that it should be called that. And I think Zhen is right: if we did want to go down that route, perhaps a better name would be what he suggests (or, worse yet, co-(fibre product)). I think all of these are bad. $\endgroup$ – Bruno Joyal Nov 16 '13 at 13:32
  • 2
    $\begingroup$ @AlexYoucis And you don't need to bring up schemes to justify the name "fibre product". Already in the category of sets, it is worthy of this name! :) $\endgroup$ – Bruno Joyal Nov 16 '13 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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