Hi I started studying Greub's multilinear algebra book and I found something very strange when he defines the tensor product of two vector spaces: He defines:
[...] Let $E$ and $F$ be vector spaces and let $\otimes$ be a bilinear mapping from $E\times F$ into a vector space $T$. We shall say that $\otimes$ has the universal property, if satisfies the following conditions:
$\otimes_1$: The vectors $x\otimes y$ ($x\in E$, $y\in F$) generate $T$, or equivalently $\textrm{img}(\otimes)=T$.
[...]
The problem is the equivalence: $$T=\textrm{span}\{x\otimes y: x\in E, y\in F\}\Leftrightarrow \textrm{img}(\otimes)=T. $$ Since $$\textrm{span}\{x\otimes y: x\in E, y\in F\}=\textrm{span}\{\textrm{im}(\otimes)\}$$ he is saying $T=\textrm{span}\{\textrm{im}(\otimes)\}$ if and only if $T=\textrm{img}(\otimes)$, in other words, he means $$\textrm{im}(\otimes)=\textrm{span}\{\textrm{im}(\otimes)\},$$ that sounds strange for me. Can anyone clarifies this?
It looks like the author is defining the tensor product as $(T,\otimes)$ with certain properties, and will ultimately say "$T$ is unique up to isomorphism, and we call it $E\otimes F$."
You're right that the set theoretic image of the function $\otimes$ is not usually a vector space, it is just a subset of $E\otimes F$. The image is usually not even closed under addition. It does span $E\otimes F$, though.
I guess it hinges on the author's notation. I would have preferred to write $\langle im(\otimes)\rangle=T$ if I wanted to express the span. The author may have adopted some convention allowing them to omit the spanning angle brackets.