# Question about the definition of tensor product of homomorphisms

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?

• In nyour definition of the ten,sor product of homomorphisms, I think it should be $f\otimes g$ everywhere. Commented Jul 22, 2020 at 13:20

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 $$A\otimes B\to A'\otimes B'$$.

• Hi @Angina Seng, thank you very much for your answer. Regarding your last paragraph, the technical details are they the same covered by Lee Mosher's answer here? math.stackexchange.com/q/1402612
– Zest
Commented Jul 22, 2020 at 14:53
• Or to put it differently: Would you mind explaining your last paragraph regarding the induced map? Does that simply follow from the homomorphism theorem?
– Zest
Commented Jul 22, 2020 at 14:59
• Let's say you have a homomorphism of Abelian groups $\phi:G\to H$ and subgroups $G_1$ and $H_1$ of $G$ and $H$. Then $\phi$ induces a homomorphism $G/G_1\to H/H_1$ sending the coset $a+G_1$ to $\phi(a)+H_1$ as long as $\phi(G_1)\subseteq H_1$. To check that, all you need is $\phi(a)\in H_1$ where $a$ runs through a set of generators for $G_1$. Commented Jul 22, 2020 at 15:03
• Ok, this makes sense. But my issue here is that we already assume that $\phi\colon G\to H$ is a homomorphism, but i was struggling to prove that my map from the set of formal generators $\{a\otimes b\}$ to the set of formal generators $\{f(a)\otimes f(b)\}$ is a homomorphism in the first place.
– Zest
Commented Jul 22, 2020 at 22:44
• Recall we are first considering the free Abelian group generated by the $a\otimes b$. Commented Jul 23, 2020 at 3:20