# Does linearization of categories reflect isomorphism?

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.

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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
If it fails for some nontrivial ring, then it fails for some field. – Tom Goodwillie Jun 21 '10 at 12:42
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

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 $C$.

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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
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
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

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...

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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
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

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 $\mathscr{C})$.

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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
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