Extending a bilinear map $f:U\times V\rightarrow W$ to a unique homomorphism $f_1 :A\rightarrow W$.

Question

In chapter 4.7 of Algebra Vol. 2 the auther, P.M. Cohn, considers a commutative ring $K$ and $K$-modules $U,V,W$, and defines $A$ as the free $K$-module formed on the set $U\times V$. He then writes

Let $f :U\times V\rightarrow W$ be any bilinear mapping; regarded as a set mapping, i.e. ignoring bilinearity, it may be extended to a unique homomorphism $f_1 :A\rightarrow W$, because $A$ is free on the elements $(u,v)$.

What I understand from the above is that we define $f_1(\alpha(u,v)):=\alpha f(u,v)$ and extend by linearity. This, I believe, makes $f_1$ a well-defined homomorphism, since $f$ is well-defined. The uniqueness arises from the fact that any homomorphism defined on the generators of $A$ is uniquely determined by its values on the generators. Is this correct, or have I misunderstood something? Also, why do we need to ignore bilinearity of $f$ in the definition of $f_1$?. Edit: do we ignore bilinearity because $f_1$ takes a single argument of the form $\alpha(u,v)\in A$, and bilinearity requires two arguments?

Another question: he later writes that we have $[(\alpha u, v)-\alpha(u, v)] f_1=(\alpha u, v) f-\alpha[(u, v) f]=0$. Is there any particular reason why he writes $(u,v)f_1$ rather than $f_1(u,v)$?

Answer 1

Note that when we define $f_1$ as a bilinear map, we are considering $U\times V$ as a product of vector spaces; that is, we are viewing it as a set with structure. So that when we see $(u_1+u_2,v)$ as an element, we think of it as "the result of adding two elements of $U$ In the first entry", etc.

However, $A$ is the free vector space on the set $U\times V$. Here, the elements of $U\times V$ do not carry any structure, in so far as $A$ is concerned. So for example, the element $(2u,v)$ has no relation whatsoever to the element $(u,v)$ of $U\times V$; the elements of $U\times V$ are simply labels we slap on the basis elements of $A$.

This is similar to the way in which we may take a free vector space on the set $\{v_0,v_1,v_2\}$, and even though we have used the elements of $\mathbb{Z}/3\mathbb{Z}$ as indices for the vectors in this basis, the structure of $\mathbb{Z}/3\mathbb{Z}$ is completely irrelevant to the fact that we have a vector space with basis $\{v_0,v_1,v_2\}$. We don't care that the index of $v_2$ is twice the index of $v_1$... we only care that they are different indices, hence different vectors.

So when we look at $A$, we don't really care that the function $f_1$ is bilinear, because we aren't thinking of $U\times V$ as a vector space; we are thinking of it simply as a set. So we can "forget" the bilinearity of $f_1$. Similarly, above, we can define a function $f\colon\{v_0,v_1,v_2\}\to \mathbb{R}$ that sends $v_0$ to $0$, $v_1$ to $1$, and $v_2$ to $-1$; this map will be "multiplicative on the indices modulo $3$"; that is, $f(v_i)f(v_j) = f(v_{ij})$, where the index $ij$ is taken modulo $3$. And because we have a map from the basis $\{v_0,v_1,v_2\}$ to the vector space $\mathbb{R}$, we can extend it to a linear function; this function is defined without regards to the fact that $f$ was "multiplicative in the indices"; that may be relevant to other things we will do later, but it does not matter for defining the linear extension from the free vector space on $\{v_0,v_1,v_2\}$ to $\mathbb{R}$ that extends $f$.

To answer your second question: Ring theorists often use suffix functional notation when working with modules (the function appears after its argument, rather than before); this is likely why Cohn is using suffix notation.

Answer 2

We ignore bilinearity for the point you mentioned i.e., extending the morphism. But there is a reason why it is said bilinearity in the first place.

In fact $f(U \times V) \subseteq (U \otimes_K V)/Ker(f)$, precisely because of bi-linear property of $f$, since tensor product needs to be bi-linear. Basically image of $u \times v$ is an equivalence class w.r.t bi-linearity i.e., same as $[f(u \times v)] = f(u \otimes v)$ but this equivalence class is w.r.t $W$.

But for the extension of morphism, its enough to view $U \times V$ as a set with distinct elements and not equivalent classes as it generates $A$ as a free module and hence can be viewed as a set. Here w.r.t extending morphism, no need to worry about bi-linearity and can be ignored.

Basically to characterize the image as $f_1(A) \cong (U \otimes_K V) / Ker(f_1)$ we usually assume that $f_1$ is also bi-linear. So it is to characterize the image as tensor product that we need to assume bilinearity and not for extension of morphism which you mentioned.