# Does this construction verify the universal property?

• A
• Heidi
Heidi
Hi Pfs,
https://en.wikipedia.org/wiki/Free_product
I see how G * H is built but how to prove that it verifies the coproduct universal property:?

Heidi said:
Hi Pfs,
https://en.wikipedia.org/wiki/Free_product
I see how G * H is built but how to prove that it verifies the coproduct universal property:?
You basically just write it down carefully. It is a matter of notation rather than a matter of mathematics. It is constructed to fulfill the universal property.

The remark with the coproduct refers to the direction of the arrows compared to direct products and direct sums. We have these three "co-/universal" constructions: free product, direct product, and direct sum. Each of them is a mess of commuting diagrams. I always have to look it up in order to determine who chases whom. If you now add projective and injective limits then the confusion is complete!

Such questions, i.e. those which concern category theory are best looked up on nlab:
https://ncatlab.org/nlab/show/free+product+of+groups

topsquark
I see that you got a "like" but i still do not understand. Sorry...
the construction is made to fulfill a condition but if i am not confident how to see that it works
How to build the unique morphisme from M * N toward the targer?

Heidi said:
I see that you got a "like" but i still do not understand. Sorry...
the construction is made to fulfill a condition but if i am not confident how to see that it works
How to build the unique morphisme from M * N toward the targer?
Why don't you try what I have said? Write down all morphisms that you have, name the group and the index set, and note what must be shown. I see no sense in rewriting the definition. So in order to communicate, we need a common language, not just M*N. Have you read the nLab link?

Which homomorphism is the problem?

Maybe this diagram will help to see why i have problem.
In my mind the universal property of group coproduct did not need amalgamated free product
I had no H only G1,G2,Q and G1 * G2 (not over H) and the injections q1,q2 and the arrows f1 , f2
and tried to understand why the construction with words, concatenation could help to
build the dashed group homomorphism f.
Have you another universal property depending on H?

Heidi said:
Maybe this diagram will help to see why i have problem.
In my mind the universal property of group coproduct did not need amalgamated free product
I had no H only G1,G2,Q and G1 * G2 (not over H) and the injections q1,q2 and the arrows f1 , f2
and tried to understand why the construction with words, concatenation could help to
build the dashed group homomorphism f.
Have you another universal property depending on H?
Ok, this simply means ##H=\{1\}.## The elements of ##G=G_1\ast G_2## are words over the alphabet of elements from ##G_1## and ##G_2.## Say we have ##abcxyacbza\in G## with ##a,b,c\in G_1## and ##x,y,z\in G_2.## Then we have a homomorphism ##f_1\, : \,G_1\longrightarrow Q## that maps ##a,b,c,## say onto ##\overline{a},\overline{b},\overline{c}## and the same for ##f_2\, : \,G_2\longrightarrow Q## that maps ##x,y,z## onto ##x',y',z'.## The dashed group homomorphism ##f## maps then ##abcxyacbza## onto ##f_1(a)f_1(b)f_1(c)f_2(x)f_2(y)f_1(a)f_1(c)f_1(b)f_2(z)f_1(a)= \overline{a}\overline{b}\overline{c}x'y'\overline{a}\overline{c}\overline{b}z'\overline{a}.##

Heidi
Thanks.

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