Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a category $C$ and a commutative ring $R$, denote by $RC$ the $R$-linearization: this is the category enriched over $R$-modules which has the same objects as $C$, but the morphism module between two objects $x$ and $y$ is the free $R$-module on $Hom_C(x,y)$. Thus in $RC$ we allow arbitrary $R$-linear combinations of morphisms from the original category $C$.

Question: if two objects in $x$, $y \in C$ are isomorphic in $RC$, are they already isomorphic in $C$?

I do not know the answer to this question for any nontrivial ring $R$, but I'm particularly interested in $R=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$.

What's obviously not true is that every isomorphism in $RC$ comes from an isomorphism in $C$ (take $-id_x$). (Thus the word "isomorphism" in the title refers to a relation on objects rather than to a property of morphisms.)

Of course, it is enough to consider categories $C$ with two objects $x$, $y$, but we cannot assume that $C$ is finite.

It's fairly elementary to see that if $x$ and $y$ are isomorphic in $RC$ then in $C$, $x$ is a retract of $y$ and vice versa, but the latter does in general not imply that $x \cong y$.

A more catchy way of phrasing this problem is: can we always classify objects in a category up to isomorphism by means of functors taking values in $R$-linear categories? (The inclusion $C \to RC$ is the universal such functor.)

Edit: A lot of people have posted an "answer" that wasn't, and deleted it, so here's something that will not work, to save others going down the same road. I said that we cannot assume that the category is finite; in fact, it must be infinite. Here is an elementary argument:

Since $x$ and $y$ are mutual retracts, there are maps $f,\;f'\colon x \to y$ and $g,\;g'\colon y \to x$ with $fg=id$ and $g'f'=id$. Consider the powers of $fg' \in End(y)$. If $End(y)$ is finite then $(fg')^n = (fg')^m$ for some $m \neq n$; since $fg'$ has a right inverse (viz, $f'g$), we must have that $(fg')^n=id$ for some $n>0$. So we see that $g'$ has not only a right inverse ($f'$) but also a left inverse: $(fg')^{n-1}f$. So they are the same and $g'$ is already an isomorphism.

share|improve this question
    
I assume that you do not want $0=1$ in $R$? –  Theo Johnson-Freyd Mar 9 '10 at 4:18
    
My guess is that the answer to your question is going to turn on something like whether $R$ has zero divisors. –  Theo Johnson-Freyd Mar 9 '10 at 4:22
    
@Theo: $0 \neq 1$ is what I mean by nontrivial... My feeling is that zero-divisors do not play a role, but I'd be curious if you could elaborate on that guess. –  Tilman Mar 9 '10 at 8:01
5  
If it fails for some nontrivial ring, then it fails for some field. –  Tom Goodwillie Jun 21 '10 at 12:42
3  
Just a comment on terminology. The inclusion functor L : C --> RC reflects isos in the usual sense. That is, if L(g) is an iso iff g is an iso. I think the hard thing to show here is actually the L creates isomorphisms. That is, for an iso L(x) --> y, there exists y', i' such that L(y') = y and i' : x --> y' is an iso. Of course, L is identity on objects, so the existence of i' is the interesting part. –  Aleks Kissinger Aug 16 '10 at 12:41
show 4 more comments

3 Answers

In the interest of having an undeleted answer, here is a small result. Let $x, y$ be objects and $f, g : x \to y$ and $u, v : y \to x$ be morphisms in $C$, and let

$$F = af + bg, G = cu + dv$$

be two morphisms in $RC$, where $a, b, c, d \in R$. If $FG = \text{id}_y, GF = \text{id}_x$, then WLOG $fu = \text{id}_y$ and also some term in $GF$ must equal $\text{id}_x$. If we want $x, y$ to be non-isomorphic, then $f$ cannot have a left inverse and $u$ cannot have a right inverse, so it must be the case that $vg = \text{id}_x$ and moreover no other composition of morphisms except $fu$ or $vg$ can be an identity.

It follows that $ac = bd = 1$, hence $a, b, c, d$ are all units, so none of the four terms in $FG$ or in $GF$ vanish. Thus the only way for all of the non-identity terms to cancel is if $gu = fv = gv$ and $ug = vf = vg$. But this implies

$$gug = fvg = f = gvg = g$$

and symmetrically $u = v$, so in fact $x, y$ must be isomorphic in $C$. Next on the list is linear combinations of three morphisms...

share|improve this answer
    
And you will probably get it for three morphisms. But I didn't see a very clean way of arguing this myself. One curious thing about the line of argument you have presented (and that I was pursuing myself) is that it really has little to do with linear algebra, and rather more with finite combinatorics (to put it vaguely!). –  Todd Trimble Mar 20 '11 at 23:10
2  
I ran a computer search a while ago to find a counterexample, but the complexity is very exponential, so I didn't get far, but I think there's no counterexample up to 3x5 morphisms. –  Tilman Mar 21 '11 at 1:41
add comment

In this answere I (try to) present the problem as a Algebraic Geometry one:

consider the category $\mathscr{C} $ with two objects $X, Y$ and

$\mathscr{C}(X, Y)$={$r_1,\ s'_1,\ r_2,\ s'_2$} ; $\mathscr{C}(Y, X)$={$s_1,\ r'_1,\ s_2,\ r'_2$} ; $\mathscr{C}(X, X)$={$1_X, e_X$} ; $\mathscr{C}(Y, Y)$={$1_Y, e_Y$} where $e_X,\ e_Y$ are idempotent, and any composition of a morphism by a a idempotent not alter the morphism, and $ 1_Y= r_1\circ s_1= r_2\circ s_2$, $ 1_X= r'_1\circ s'_1= r'_2\circ s'_2$, all other compositions give the (no identity) idempotent. Suppose that $R$ is a commutative ring and in $R\mathscr{C} $ consider the morphims $A:= a_1\cdot r_1 + b'_1\cdot s'_1 + a_2\cdot r_2 + b'_2\cdot s'_2: X\to Y$ and

$B:= b_1\cdot s_1 + a'_1\cdot r'_1 + b_2\cdot s_2 + a'_2\cdot r'_2: Y\to X$.

Let $\alpha :=a_1+b'_1+ a_2+ b'_2$, $\beta :=b_1+a'_1+ b_2+ a'_2$,

Then we have $B\circ A=1_X$ iff:

1) $ a'_1\cdot b'_1+ a'_2\cdot b'_2=1$ and

2) $\beta \cdot a_1+ (\beta - a'_1)\cdot b'_1+ \beta \cdot a_2+(\beta -a'_2)b'_2=0$ i.e. $\beta \cdot \alpha = a'_1\cdot b'_1+ a'_2\cdot b'_2 $

similarly we have $A\circ B=1_Y$ iff:

1') $ a_1\cdot b_1+ a_2\cdot b_2=1$ and

2') $\alpha \cdot \beta = b_1\cdot a_1+ b_2\cdot a_2 $

all these equations are equivalent to the system of three equations:

$ a'_1\cdot b'_1+ a'_2\cdot b'_2=1,\ a_1\cdot b_1+ a_2\cdot b_2=1,\ \alpha \cdot \beta = 1$

thinking these in $\mathbb{C}[ a_1, b'_1, a_2, b'_2 , b_1, a'_1, b_2,a'_2] $ these represent three varieties on $\mathbb{C}^8 $

If these varieties have an a intersections then $X, Y$ are isomorphic in $\mathbb{C}\mathscr{C} $ (but aren't isomorphic in $\mathscr{C}) $.

share|improve this answer
2  
Buschi, while we all make mistakes, I have noticed a frequent tendency on your part not to think through your answers carefully before posting. Observe that $s_2 \circ r_1 \circ s_1 = s_2 \circ 1_Y = s_2$, and $s_2 \circ r_1 \circ s_1 = e_X \circ s_1 = s_1$ according to your definitions. So $s_1 = s_2$. By similar calculations, all the morphisms $Y \to X$ are equated, as are all morphisms $X \to Y$. Please do not post any more answers to this question unless you are reasonably certain that you are not overlooking something. –  Todd Trimble Mar 22 '11 at 17:42
    
Thank you Trimble, only I love learn math (and sometime do mine little best), and I see that MAth is mother of humilty as well as knowledge, I hope improve both. –  Buschi Sergio Mar 22 '11 at 20:08
2  
Okay. I guess from your response that Buschi is your last name? I will use Sergio in the future if you prefer that. –  Todd Trimble Mar 22 '11 at 20:13
    
See my edit. Finite categories won't work. –  Tilman Mar 22 '11 at 21:17
    
Yes , Sergio is my name, Buschi is the surname. Call my Sergio, it's ok –  Buschi Sergio Mar 22 '11 at 21:53
add comment

Hi Tilman. I believe I proved that (in your language) linearization reflects isomorphism. The following is a sketch. I will send you a more detailed version. The general case may be reduced to the case of prime fields $F_p$ and certain categories $C$ with fixed objects $x$ and $y$ and morphisms $f_1,\dots,f_m\colon x\to y$ and $g_1,\dots,g_n\colon y\to x$ subject to relations which correspond to the fact that $u=f_1+\dots+f_m$ and $u^{-1}=g_1+\dots+g_n$ are mutually inverse in the $F_p$-linearization. Apart from trivial cases, we may reindex these generators such that $f_1g_1 = 1_y$ and $g_nf_m=1_x$, while the other summands in the expansion of $uu^{-1}$ and $u^{-1}u$, respectively, fall into equivalence classes whose size is a multiple of $p$. It is then possible to derive a sequence of pairs $(i_1,j_1),(i_2,j_2),\dots,(i_k,j_k)$ such that $f_{i_r}g_{j_r} = f_{i_{r+1}}g_{j_{r+1}}$ for $r=2,3,\dots,k-1$ and $g_{j_r}f_{i_r}=g_{j_{r+1}}f_{i_{r+1}}$ for $r=1,3,\dots,k-2$. Then $f_{i_1}g_{j_2}f_{i_3}g_{j_4}\dots f_{i_k}$ and $g_{j_k}f_{i_{k-1}}g_{j_{k-2}}f_{i_{k-3}}\dots g_{j_1}$ are mutual inverses of $C$.

share|improve this answer
4  
I read your note and I'm convinced. Would you care to share your solution with the community by posting a link to it? –  Tilman Feb 13 '12 at 7:56
2  
A new version of this note can be downloaded from idmp.uni-hannover.de/downloads_wd.html - Winfried –  Winfried Mar 4 '12 at 8:13
    
I didn't notice that this problem has been finally solved! Great! The current proof doesn't use any reductions anymore mentioned in the answer. –  Martin Brandenburg May 15 '12 at 15:34
1  
Thank goodness this problem has been put to rest at long last! That's wonderful. Wish I had noticed earlier. –  Todd Trimble Apr 7 '13 at 2:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.