I have a problem connecting the definition of a coproduct with its often mentionend universal property. Let's start with the definition (just for two objects):

Let $A_1$ and $A_2$ be objects of a category $\mathcal{C}$. A coproduct of $A_1$ and $A_2$ in $\mathcal{C}$ is a triple $(A,p_1,p_2)$ where $A \in ob(\mathcal{C})$ and $p_i \in hom_\mathcal{C}(A_i,A)$ such that if $B$ is any object in $\mathcal{C}$ and $f_i \in hom_\mathcal{C}(A_i, B), i=1,2$, then there exists a unique $f \in hom_\mathcal{C}(A,B)$ such that the diagrams are commutative ($f_i = p_if$).

My apologies, I can't draw the diagram but it's pretty simple with the equation and I assume well known. While this already sounds like a universal property to me, referred to as the universal property of the coproduct is something else:

$$hom(A_1 \amalg A_2, B) \cong hom(A_1,B) \prod hom(A_2,B)$$

Now I wonder how to connect these two, e.g. how to derive this statement from the definition. It comes out of the blue for me, and I didn't came across an explanation nor a proof so far. Also often I see $\mathrm{Hom}$ instead of $hom$ - that is the same, isn't it?

  • 1
    $\begingroup$ In your displayed formula, when you wrote $hom(A\times B, C)$, did you mean for the $\times$ sign to be a coproduct sign $\amalg$ (\amalg) or $\oplus$ (\oplus)? $\endgroup$ – MJD Oct 25 '14 at 16:25
  • $\begingroup$ Thanks, since I'm not familiar with this second part, I can't say for sure, but I think it should be the coproduct sign. My mistake. Often it's also written in the middle, but that should be the same. $\endgroup$ – cbb Oct 25 '14 at 16:27
  • $\begingroup$ Can you tell us where you saw this universal property mentioned? Someone here might have the same book and be able to explain in detail what was meant. $\endgroup$ – MJD Oct 25 '14 at 16:28
  • 2
    $\begingroup$ I think you also mean to say $\operatorname{hom}(A,C)\times \operatorname{hom}(B,C)$. Whatever the case, think about how, when these hom-classes are hom-sets, the universal property tells you how to take two morphisms $f:A\rightarrow C$ and $g:B\rightarrow C$ and construct the map $f\amalg g:A\amalg B \rightarrow C$. $\endgroup$ – Hayden Oct 25 '14 at 16:28
  • 1
    $\begingroup$ Sorry for beeing so incorrect, you are absolutely right. Book is a good question - my problem is that none of my sources include both, but at the Wikipedia entry it's mentionend (generalized for n objects): link The statement will be important to show that the tensor product shares the same universal property in a special case: link $\endgroup$ – cbb Oct 25 '14 at 16:33

Let $x$ and $y$ be objects of a category $C$. Their coproduct is an object $x \sqcup y$ together with morphisms $x \to x \sqcup y$ and $y \to x \sqcup y$, satisfying the universal property that given any object $z$ with maps $x \to z$ and $y \to z$, there exists a unique morphism $x \sqcup y \to z$ through which the given maps factor. Another way to say this is that the data of

  • an element of the set $\mathrm{Hom}(x \sqcup y, z)$

is the same as the data of

  • an element of the set $\mathrm{Hom}(x, z)$,
  • and an element of the set $\mathrm{Hom}(y, z)$.

Of course, this is the same as the data of

  • an element of the product of the sets $\mathrm{Hom}(x, z) \times \mathrm{Hom}(y, z)$.
  • $\begingroup$ Thank you, that helped a lot! It is now clear, that for any element in $\mathrm{Hom}(x, z) \times \mathrm{Hom}(y, z)$ one can find an element in $\mathrm{Hom}(x \sqcup y, z)$ through the existence of a unique morphism. But why is the other way round true? Couldn't theoretically two different elements in $\mathrm{Hom}(x, z) \times \mathrm{Hom}(y, z)$ have the same morphism in $\mathrm{Hom}(x \sqcup y, z)$? Or even more, maybe there exists a $g \in \mathrm{Hom}(x \sqcup y, z)$ for which no fitting morphisms from $x$ and $y$ to $z$ exists? $\endgroup$ – cbb Oct 26 '14 at 10:23
  • $\begingroup$ Given a morphism $x \sqcup y \to z$, one gets morphisms $x \to x \sqcup y \to z$ and $y \to x \sqcup y \to z$ by precomposition. Note that morphisms $x \to x \sqcup y$ and $y \to x \sqcup y$ are part of the definition of the coproduct. Does this answer your question? $\endgroup$ – user314 Oct 26 '14 at 11:15
  • $\begingroup$ Yes indeed, thank you! $\endgroup$ – cbb Oct 26 '14 at 11:22

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.