Mathematics Asked by Zest on November 24, 2021
I am working on an exercise about tensor products. We introduced them as the quotient space given by the following definition:
For two Abelian groups $A$ and $B$ we define their tensor product
$Aotimes B$ as the quotient of the free Abelian group on the set of
formal generators ${a otimes b mid a in A; b in B}$ by the
subgroup generated by elements of the form $$a_1 otimes b + a_2
otimes b − (a_1 + a_2) otimes b$$ and $$aotimes b_1 +aotimes b_2
−aotimes(b_1 +b_2).$$ By abuse of notation we write $aotimes b$ for
the corresponding element in the quotient $A otimes B.$
Now what i am working on is:
Two homomorphisms $fcolon Ato A’$ and $gcolon Bto B’$ induce a homomorphism $$fotimes bcolon Aotimes B to A’otimes B’ text{with} fotimes g(aotimes b) = f(a)otimes g(b)$$
So the proposed solution says that the prescription of this map is a well defined homomorphism from the set of formal generators ${a otimes b mid a in A; b in B}$ to the quotient $A’otimes B’$.
However, i can’t see why this is supposed to be the case.
In order to verify $fotimes g$ is a (well defined) homomorphism i need to show that
$$fotimes gcolon {a otimes b mid a in A; b in B}to A’otimes B’, (aotimes b)+(a’otimes b’) mapsto [f(a)otimes g(b)]+[f(a’)otimes g(b’)]$$
where $[cdot]$ denotes an equivalence class in the quotient $A’otimes B’$.
Now what i’ve tried is to work out whether the map between the set of formal generators
$$widetilde{fotimes g}colon {a otimes b mid a in A; b in B}to {f(a) otimes f(b) mid f(a) in A’; g(b) in B’}$$
is a homomorphism, because that would imply that the composition $fotimes g = pcircwidetilde{fotimes g}$ with the projection map $p$ would be a homomorphism and i would be done (after checking well-definedness).
But the issue i am having is that i do not know how to prove that $widetilde{fotimes g}$ itself is a homomorphism, given the definition above.
Please note: I would like to solve this without any usage of the universal property and bilinear-maps. Is it possible to solve it just by the definition via the quotient from above?
Where does $fotimes g$ send your relation $$a_1otimes b+a_2otimes b-(a_1+a_2)otimes b?$$ It sends it to $$f(a_1)otimes g(b)+f(a_2)otimes g(b)-f(a_1+a_2)otimes g(b)$$ which equals $$f(a_1)otimes g(b)+f(a_2)otimes g(b)-(f(a_1)+f(a_2))otimes g(b)$$ and that is precisely one of the relations in $A'otimes B'$.
The same works for all relations of the second type. Since the map $aotimes bmapsto f(a)otimes g(b)$ sends relations to relations, it induces a map $Aotimes Bto A'otimes B'$.
Answered by Angina Seng on November 24, 2021
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