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
$A\otimes B$ 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 $$a\otimes b_1 +a\otimes b_2
−a\otimes(b_1 +b_2).$$ By abuse of notation we write $a\otimes b$ for
the corresponding element in the quotient $A \otimes B.$
Now what i am working on is:
Two homomorphisms $f\colon A\to A'$ and $g\colon B\to B'$ induce a homomorphism $$f\otimes b\colon A\otimes B \to A'\otimes B'\ \ \text{with}\ \ f\otimes g(a\otimes b) = f(a)\otimes g(b)$$
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 $f\otimes g$ is a (well defined) homomorphism i need to show that
$$f\otimes g\colon \{a \otimes b \mid a \in A; b \in B\}\to A'\otimes B',\ (a\otimes b)+(a'\otimes b') \mapsto [f(a)\otimes g(b)]+[f(a')\otimes g(b')]$$
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
$$\widetilde{f\otimes g}\colon \{a \otimes b \mid a \in A; b \in B\}\to \{f(a) \otimes f(b) \mid f(a) \in A'; g(b) \in B'\}$$
is a homomorphism, because that would imply that the composition $f\otimes g = p\circ\widetilde{f\otimes g}$ 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 $\widetilde{f\otimes g}$ 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?
Best Answer
Where does $f\otimes g$ send your relation $$a_1\otimes b+a_2\otimes b-(a_1+a_2)\otimes b?$$ 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 $a\otimes b\mapsto f(a)\otimes g(b)$ sends relations to relations, it induces a map $A\otimes B\to A'\otimes B'$.