Yesterday I was working on an exercise-sheet, given 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 $A \otimes B.$
Today I wanted to verify the following:
Two homomorphisms $f\colon A\to A'$ and $g\colon B\to B'$ induce a homomorphism $$f\otimes b\colon A\otimes B \to A'\otimes B'\ \ \text{with}\ \ f\otimes g(a\otimes b) = f(a)\otimes f(b)$$
I know that I am supposed to verify $f\otimes g$ is well defined and $f$ and $g$ indeed induce $f\otimes g$. But I wasn't too confident in how to do this exactly. I was wondering, whether my initial idea about picturing the diagram as below is how to approach problems like this in general.
$\hskip2in$
Where $\{a^n\}$ denotes the set of formal generators of $A$ and $i_*$ are the inclusion maps. That would somehow reduce the problem to the universal property of the quotient space
$\hskip2in$
Would my approach be correct? Did I "think about it in the right way"? Or is there another "canonical" way to this problem?
This exercise might be trivial to most people, but I do not have too much experience in these kind of problems and very little practise with commutative diagrams or diagram chasing in general. I would really love to learn the general approach to problems like this.
It is preferable to forget how the tensor product is constructed, and to remember its universal property: any $\mathbb{Z}$-bilinear map $\varphi:A\times B\to C$ (where $C$ is an abelian group) induces a unique group morphism $u:A\otimes B\to C$ such that $u(a\otimes b)=\varphi(a,b)$ for all $(a,b)\in A\times B$.
Now, I let you check that $\varphi: (a,b)\in A\times B\mapsto f(a)\otimes g(b)\in A'\otimes B'$ is $\mathbb{Z}$-bilinear...