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Defining the Tor-Functor with a classical flavour | Rene Recktenwald Defining the Tor-Functor with a classical flavour

Rene Recktenwald

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Defining the Tor-Functor with a classical flavour

$\DeclareMathOperator{\Tor}{Tor}$

I

Let $F:\Aa\to \Bb$ be right exact functor between sufficiently nice abelian categories. As everyone remembers from the tender days of childhood we can define the left derived functors $L^iF$ of $F$ in the following way: For $A\in \Aa$ let $P^\bu\to A\to 0$ be a projective resolution. Then $L^iF(A):=H^i(F(P^\bu))$. One has to ask oneself why this is well-defined (up to unique isomorphisms) and why it is a functor. Let me just sketch the argument here.

Let $A\in \Aa$ be an object. Let $P^0\to A\to 0$ be an epimorphism where $P^0$ is projective. Let $K^0$ denote the kernel, consider $P^1\to K^0\to 0$. We can compose this to $P^1\to P^0\to A$. Now we repeat this process to build a projective resolution of $A$: At every point we take the kernel (note that it does not matter whether we take the kernel of $P^i\to P^{i-1}$ or $P^i\to K^{i-1}$), and ‘resolve’ the kernel with a projective object. In particular the resulting complex $P^i\to P^{i-1}\to \ldots to A\to 0$ is exact. What we have to show is, that if $P^\bu\to A$ and $Q^\bu\to A$ are two projective resolutions then there is a natural isomorphism $H^i(F(P^\bu))\cong H^i(F(Q^\bu))$.

However it is actually quite natural to simply show that a morphism $f:A\to B$ induces a morphism $L^i(F(A))\to L^i(F(B))$, and then for $f=\id_A$ we immediately get the desired isomorphism.

Onwards! Let $f:A\to B$ be a morphism, and let $P^\bu\to A$ and $Q^\bu\to B$ be projective resolutions. Via composition we have a morphism $P^0\to B$ and by the universal property of projective objects (in this case we apply it for $P^0$) there is a morphism $P^0\to Q^0$, such that the obvious square commutes.

\[\begin{CD} P^1 @>>>P^0 @>>> A@>>>0\\ @. @VVf^0V @VVfV\\ Q^1 @>>>Q^0 @>>> B @>>>0 \end{CD}\]

For the next step note that when we consider $P^1\to P^0\to Q^0\to B^0$ everything gets send to zero, since $P^1\to P^0\to A$ is the zero morphism. This induces $P^1\to \ker(Q^0\to B)$. But since $Q^1\to \ker(Q^0\to B)$ is an epimorphism by construction we get $P^1\to Q^1$. Again this is the universal property of $P^1$.

We again simply continue in this fashion to get a morphism of complexes $P^\bu\to Q^\bu$. Now it is clear that $F$ extends to a functor of complexes, i.e. there is a morphism of complexes $F(P^\bu)\to F(Q^\bu)$. And it is true, in general, that a morphism of complexes $C^\bu\to D^\bu$ induces natural morphisms $H^i(C^\bu)\to H^i(D^\bu)$, this is a bit tedious, but not hard to show.

II

Wait we have not even used that $F$ is right exact‽ Yeah we haven’t. Up to this point you could play the same game with any covariant additive functor. However

Proposition $L^0F$ is naturally equivalent to $F$ if and only if $F$ is right exact.

Proof If $F$ is right exact, then $F(P^1)\to F(P^0)$ is surjective, with cokernel $F(A)$. Since cokernels are essentially unique it follows that $L^0F\to F$ is a natural isomorphism. On the other hand let $L^0F$ and $F$ be naturally isomorphic, and let $A\to B\to 0$ be an epimorphism, where we choose projective resolutions $P^\bu\to A$ and $Q^\bu\to B$ as before. By definition of $H^0$ the natural morphisms $F(Q^0)\to L^0F(B)$ is an epimorphism. By assumption it is $L^0F(A)\cong F(A)$ and $L^0F(B)\cong B$. Now observe that by the universal property of $Q^0$ there is a morphism $Q^0\to A$, which gives rise to a morphism $F(Q^0)\to F(A)$. But that means that we could lift an epimorphism along $F(f)$, so it must itself be an epimorphism! (This is a simple lemma that holds in any category.)

III

What happens if we try to mimic the same construction in some category of sheaves with flat sheaves, instead of projectives? Well, the first part doesn’t work. Given a morphism $A\to B$ and flat resolutions of $A$ and $B$, there is no reason to expect morphisms between the resolutions. Indeed we have only used the fact that the functor is right exact at the very end, everything before were possible purely by the might of projective objects. The fact that flat sheaves behave really well with tensor products does not save us; we do not even reach a point where there is a morphism to apply the tensor product to!

There is even a simple example where no suitable map exists. We consider the category of $\Z$-modules, i.e. abelian groups. Consider the identity $\Z/p\Z\to \Z/p\Z$. Then $Z\to\Z/p\Z$ is a flat resolution, of course. But so is $\Z_{(p)}\to \Z/p\Z$. $\Z_{(p)}$ is flat since it is a localization. However there is no morphism $Z_{(p)}\to \Z$.

IV

In order to remify this defect we need a stronger requirement than simply, ‘There is some flat sheaf $F$ with an epimorphism $F\to A$’. In 1981 Enochs conjectured (link) that they exist, and 20 years later Bican, Bashir and Enochs were able to prove it (link); it’s just 6 pages, go have a look.

So what are flat covers, and how can they help us?

For a class of objects $\Dd$ in a category $\Cc$ then a morphism $f:D\to A$ with $D\in \Dd$ is a $\Dd$-pre-cover if for any $D’\in \Dd$ and any morphism $D’\to A$ there is a morphism $D’\to D$ making the obvious triangle commute. If additionally the only endomorphisms of $D$ commuting with $f$ are automorphisms, then we call it a $\Dd$-cover.

Before we continue we should toy around with this definition and projective objects. Let $f:P\to A$ be an epimorphism with $P$ projective. Let $P’$ be any projective object and $P’\to A$ be any morphism. Then by the universal property of $P’$ there is a morphism $P’\to P$ making the triangle commute. Now let $g:P\to A$ be any morphism, can we only extend this triangle with automorphisms? No. Consider the category of $k$ vector spaces, let $A=k$, $P=k^2$ and $f=g$ be the projection to the first coordinate. Then we can do whatever we want on the second coordinate, and still get a triangle. We see that being a pre-cover is relatively straight forward for projectives, but being a cover is harder. But that should not be surprising. After all covers are defined via an universal property, so as always there is the following

Proposition $\Dd$-covers are unique up to non-unique isomorphisms.

Proof Let $f_1,f_2:D_1,D_2\to A$ be $\Dd$-covers. Then there is $g_2:D_1\to D_2$ such that $f_2g_2=f_1$. Similarly we find $g_1$ with $f_1g_1=f_2$. Then $f_1g_1g_2=f_2g_2=f_1$. And by the property that distinguished pre-covers from covers we see that $g_1g_2$ is an automorphism.

V

Now we mimic the construction of projective resolutions, but at each step we take a flat pre-cover, which luckily we know always exists. Let $F^\bu\to A$ and $G^\bu\to B$ be such resolutions and let $f:A\to B$ be any morphism. Then we have $F^0\to B$ which induces $F^0\to G^0$ because $G^0\to B$ is a pre-cover. Next we consider $F^1\to F^0\to G^0\to B$. By commutativity this is the zero morphism and we obtain $F^1\to \ker(G^0\to B)$, but $G^1\to \ker(G^0\to B)$ is an epimorphism by construction. The universal property of $G^1$ induces $F^1\to G^1$. Again this is just the same argument as for the projective case. The only subtle difference is that now we are using the universal properties of the resolution of $B$, instead of the resolution of $A$.

\[\begin{CD} F^1 @>>>F^0 @>>> A@>>>0\\ @. @VVf^0V @VVfV\\ G^1 @>>>G^0 @>>> B @>>>0 \end{CD}\]

Therefore if we always choose flat pre-covers when resolving a sheaf, we could define the $\Tor$-functor in a classical way.