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In this article, there are two proposition as the following:

1.(Proposition.) For every bilinear map $f:V\times W\to U$ there is a unique linear map $h:V\otimes W\to U$ such that $hg=f$, where $g$ is the bilinear map from $V\times W$ to $V\otimes W$ that takes $(v,w)$ to $v\otimes w$.

2.(Lemma.) Let $U$ and $V$ be vector spaces, and let $b:U\times V\to X$ be a bilinear map from $U\times V$ to a vector space $X$. Suppose that for every bilinear map $f$ defined on $U\times V$ there is a unique linear map $c$ defined on $X$ such that $f=cb$. Then there is an isomorphism $i:X\to U\otimes V$ such that $u\otimes v=ib(u,v)$ for every $(u,v)$ in $U\otimes V$.

The author says in the remark that "the point of the lemma is that any bilinear map $b:U\times V\to X$ satisfying the universal property is isomorphic to the map $g:U\times V\to U\otimes V$ in an obvious sense."

As I understand, when people talk about isomorphism, there should be two structures. Here comes my question:

What does "isomorphic" mean in this context? And where are the underlying two structures? (How can a map isomorphic to another map?)

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It is somewhat annoying to write these on math.SE, but it means they are related by commutative diagram (en.wikipedia.org/wiki/Commutative_diagram). –  Qiaochu Yuan Jul 15 '11 at 1:59
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@Qiaochu: why can't/don't they introduce the xymatrix package to the SE network? –  Bruno Stonek Jul 15 '11 at 2:09
    
@Bruno: I don't know whether SE has control over this. Currently math is rendered using MathJax (mathjax.org) and I can't figure out where in the documentation to find what packages they are capable of supporting. –  Qiaochu Yuan Jul 15 '11 at 2:16
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@Qiaochu: Half a year is an extremely long time in this age of computers, but according to Anton Geraschenko in this meta.MO thread there are plans but they aren't implemented, yet. –  t.b. Jul 15 '11 at 2:20
    
@Bruno: ping... –  t.b. Jul 15 '11 at 2:20

1 Answer 1

up vote 6 down vote accepted

You can talk about "isomorphic" maps without problems: that means isomorphism in a category of maps.

But we can avoid talking about categories of maps and safely delete the word "category" in this case: you have just two maps with the same source

$$ X \stackrel{b}{\longleftarrow} U \times V \stackrel{g}{\longrightarrow} U \otimes V \ . $$

For these two maps, being isomorphic means exactly the contents of the lemma: there is a third map $i : X \longrightarrow U\otimes V$, which is an isomorphism, and such that $ib = g$. (In categorical lingo, this map $i$ can be called a map from $b$ to $g$.)

In fact, the lemma could have said more: this map $i$ is unique with these properties and when you have a unique isomorphism between two guys you're the happiest mathematician in town because this is the closest thing to an equality you can usually get.

But in this case, this remark of mine adds nothing to the situation. So, coming back to your question: the remark of your paper adds nothing to the lemma.

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