Hello all, I'm working on an exercise I did a while back from Rotman's Introduction to Homological Algebra: Show ever left exact covariant functor $T : _R\text{Mod} \to \text{Ab}$ preserves pullbacks. I tried it again right now and had trouble with it, then I looked back at my original proof and realized it was quite flawed. What's the canonical way to go about proving this?
So we have $f : B \to A$ and $g : C \to A$ and the pullback is $(D, \alpha, \beta)$ where $\alpha : D \to C$ and $\beta : D \to B$ such that $f \circ \beta = g \circ \alpha$.
We can define $\varphi : B \oplus C \to A$ by $\varphi(b, c) = f(b) - g(c)$ and get the exact sequence: $$0 \to \ker \varphi \to B \oplus C \to A$$ and we can notice that $\ker \varphi = D$.
If I apply $T$ to this left exact sequence, I get: $$0 \to T(D) \to T(B \oplus C) \cong T(B) \oplus T(C) \to T(A)$$ is exact.
I also know $T(D) \cong \text{Im } T\iota = \ker T \varphi$ where $\iota : D \to B \oplus C$ is the inclusion map.
I want to show $\ker T\varphi \cong D'$ where $D'$ is the pullback of $Tf : TB \to TA$ and $Tg : TC \to TA$.
I know what $D'$ looks like, since I have a concrete construction of pullbacks in $R$-$\text{Mod}$, and since $D'$ is in the category of abelian groups, we have $R = \mathbb Z$ in this case.
@RobertCardona It really should just be a matter of writing things out. The general definition of left exact is preserving all finite (co)limits (can never remember which).
My problem is that since $T$ is an arbitrary covariant functor, I don't know exactly what $T$ does to $\varphi$ and so I don't know how to construct a map from $\ker T\varphi$ to $D'$ or backwards.
That's not the definition of left exact I have at my disposal right now :/
@RobertCardona Right, but that means it should not be something particularly mysterious going on, as you could in fact prove something more general. But I am too tired to try to recall how to do this right now
I've encountered this exact problem before trying to that if $F : _R \text{Mod} \to \text{Ab}$ is a covariant functor then $F$ preserves kernels iff $F$ is left exact which I posted here as a question. The author seems to have glossed over the exact details that caused me problems.
I also feel it should be quite obvious, but I'm just not seeing it given the definitions I have at my disposal. I keep hitting dead ends :/
I'd hate to post this as a question because I'm usually given answers that use tools I don't have at my disposal, which often make the problem more tractable. I'll keep at it a few more hours and see what I can figure out.
@MikeMiller I really don't see why the orientation $\mu : G \to \{\pm 1\}$ I defined by multiplication by $g$ is continuous in a small neighborhood of $1$, where $g$ is close to $1$.
Right. Aren't you going to tell me why it is? I guess I should start by taking a ball $B$ around $1$. Mult. by $g$ perturbs this slightly to a ball $B'$. You get an isomorphism $H_n(G|B) \to H_n(G|B')$. In fact, you get a commutative square with vertices $H_n(G|B)$, $H_n(G|B')$, $H_n(G|1)$, $H_n(G|g)$. I guess one has to fiddle with this.
You're thinking in terms of continuity of some map but this is not how local coherence of orientation in Hatcher's sense is defined. You would have to show that Hatcher's definition is equivalent to continuity of some map if you wanted to think that way.
I feel like some commutative diagram wizardry could be done here, but I get stuck each time I try to use it.
Hmm. Why can't the following be done : let $1, g$ be inside a ball $B$. Then there is a commutative triangle with vertices $H_n(G|B), H_n(G|1), H_n(G|g)$ with each arrow an isomorphism. Then it's clear that if image of $\alpha$ via $H_n(G|B) \to H_n(G|1)$ is the chosen orientation at $1$, then image of $\alpha$ via $H_n(G|B) \to H_n(G|g)$ is the same thing.
Well, if image of $\alpha$ via $H_n(G|B) \to H_n(G|1)$ is the chosen local orientation at $1$, then composing this with $H_n(G|1) \to H_n(G|g)$ is the chosen local orientation at $g$. Thus, by commutativity, image of $\alpha$ via $H_n(G|B) \to H_n(G|g)$ is the chosen local orientation at $g$
Regarding the definition of continuity, why isn't it good enough to say the limit as x -> c is the same as the f(c)? Why do we need to specify that the limit exists? If it equals f(c), then it exists, no?
@MikeMiller I can pick $B$ in a way such that $B$ is multiplication by $g$ invariant, right? orbit of $1$, maybe. I am not at all comfortable with groups having a topology.
There's a dazzlingly bright star in the east sky in this part of the world which is getting brighter every second. I am starting to wonder whether it's a meteor or something.
