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My book asserts that, for several common categories, such Vec, Grp, Top, etc., the forgetful functor is not injective on the category's class of objects, $\mathcal{Ob}$.

I'm looking for examples of this assertion for the forgetful $U_\mathbf{Vec} = U$ on Vec.

(By Vec I mean the category of vector spaces over R, with the R-linear transformations as morphisms.)

And a second, related question:

Is there a $X\in \mathcal{Ob}_\mathbf{Set}$ such that

  1. $|X| > 1$, and
  2. $U(V\;) = X$ for exactly one $V\, \in \mathcal{Ob}_\mathbf{Vec}\;\;$?

P.S. One reason for my difficulties with this question is that the common modern mathematical practice of declaring superficially different entities as "essentially the same" (as in "a vector space $V$ and its double dual $V^{**}$ are essentially the same", or "all singletons are essentially the same") makes it difficult for me to settle questions regarding injectivity, which hinge crucially on knowing when two things are different (e.g. $V, W \in \mathcal{Ob}_\mathbf{Vec}$, on the one hand, and $U(V\;), U(W\;)\in\mathcal{Ob}_\mathbf{Set}$ on the other).

Edit: added a link to the book I referenced, and changed Vect to Vec to match the book's notation.

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Dear kjo: Do you agree that $\mathbb R$ can be equipped with different $\mathbb R$-vector space structures? – Pierre-Yves Gaillard Nov 19 '11 at 16:20
@Pierre-YvesGaillard: Yes, I see this now. Thanks! – kjo Nov 19 '11 at 19:19
offtopic: What are those “\;” in your $\TeX$ code for? – beroal Nov 20 '11 at 10:21
@beroal: they are primarily tokens of my obsessive character; as a bonus, they insert a modicum of horizontal whitespace :) Question to you: what prompted you to look at the $\LaTeX$ source? – kjo Nov 20 '11 at 13:20
@kjo: I read stackexchange questions in Liferea (a feed reader). It does not run JavaScript, so I see the source of formulas. – beroal Nov 21 '11 at 20:53
up vote 4 down vote accepted

In $\bf{Vect}$ you can take one set and define different vector space structures on them. For example define a new addition $\oplus$ on the set $\mathbb{R}^2$ by setting $(a,b) \oplus (c,d) := (a + b + c+ d, b + d)$. Together with the usual scalar multiplication you get a vector space with a different vector space structure than the normal one on $\mathbb{R}^2$. Applying the forgetful functor gives you the same set.

In general you can take any bijection $\psi: \mathbb{R} \rightarrow \mathbb{R}$ and define $a \oplus b := \psi^{-1}(\psi(a) + \psi(b))$ and $a \otimes b := \psi^{-1}(\psi(a) \cdot \psi(b))$. This defines a new field structure on $\mathbb{R}$ with different units, inverse elements etc. But it will be still be a $\mathbb{R}$-vector (here $\mathbb{R}$ considered with its usal structure) space by using the above multplication for scalar multiplication, too. Generalizing this gives you a negative answer to your second question.

The general idea is that you look at a set without the "meaning" its elements have in the concrete category. You are right when you say that $V$ and $V^{**}$ are essentially the same because we have a (canonical) isomorphism in the case that $V$ is finite dimensional. One point of category theory is to make this difference between being equal and being essentially the same (meaning isomorphic) more visible.

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Thanks! I gather that $U(V\;) \neq U(V^{**})$? – kjo Nov 19 '11 at 16:53
You are right. They may be isomorphic, but $V^{**}$ contains linaer maps $V^* \rightarrow k$ and these are not elements of $V$. – Matthias Klupsch Nov 19 '11 at 16:59

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