Mathematics - 2012-03-26 (page 1 of 4)

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12:18 AM

Right hand rule with Cross-product, any easy mnemonic?

1

A: Unit vectors in Polar coordinates, show $\hat{e}_R=\frac{(x,y,z)}{r}$

$\hat{e}_R$ is the unit radial vector. This is simply $(x,y,z)$ divided by its length, $r$: $\hat{e}_R=\frac{(x,y,z)}{r}$. $\hat{e}_\theta$ is the unit vector tangent to the sphere, thus perpendicular to $(x,y,z)$, which is also perpendicular to $(0,0,1)$ since changing $\theta$ does not chan...

(I mean the minus etc thing to specify whether counter-clockwise or not)

the RHR is already an easy mnemonic imo

@anon yes but looking at the unit-vectors -- does not really ring me any bells about clockwiseness

Say, @hhh have you read the chat etiquette rules?

@AsafKaragila sorry? Which rules?

@hhh Chat rules.

In particular rule number three... :-)

@anon, you know a bit algebra. Right?

12:28 AM

A bit, yes.

@AsafKaragila I see, I was so accustomed to Unix chat where I was allowed to market my question but as you wish :) ...Unix chat is full of that kind of large marketing qs, a bit distracting -- yes.

Why, in the name of Arthur son of Uther Pendragon, when I have a free resolution of $A$, say $\to P_1\to P_0\to A\to 0$ and I apply $Hom(\bullet, M)$ to this sequence I can replace the $0\to Hom(A,M)\to\ldots$ by $0\to Hom(P_0,M)\to\ldots$??

The context is abelian groups over $\mathbb Z$.

@hhh No problem marketing your questions, just use compact link forms...

above my head, sorry.

@AsafKaragila yes not bad idea, thanks.

:-)

@anon Ah. Okay, I found it.

12:35 AM

Right hand rule -mnemonics -q, better to do things thoroughly to get things clear.

12:47 AM

Score!! I went to buy some snacks from the vending machine and since the bag got stuck against the glass the machine gave me another one, and when that got stuck it just gave my money back so I bought a third snack!

serendipity

Yes. Now if I could only finish this crap already...

1:07 AM

Ugh, and that stupid antivirus running in the background so I can hardly operate this computer anymore! >:(

@Asaf: That is because $A$ is not considered to be part of the resolution. That's why the indexing starts at $0$ rather than $1$.

@ZhenLin Yes, I understand that but the $0\to Hom(P_0,M)\to Hom(P_1,M)\to\ldots$ suggests that the induced map from $Hom(P_0,M)$ to $Hom(P_1,M)$ is injective...

It is not.

The $0$ just indicates that the complex is truncated there.

I see.

So, can you help me here with this horror show that has been haunting me for the past 8 hours or so?

Maybe. I'm not entirely awake yet...

1:13 AM

I need to prove that for abelian groups (and for that context only!) $Ext(A,M)=Ext^1_\mathbb Z(A,M)$ is independent of the free resolution of $A$.

Hmmm... I don't know whether there's any less work involved in doing that.

I know that the general case follows almost immediately from this one.

However I really get lost when all the books I found start giving out long sequences... whereas I essentially have one map and two more induced by the projective-ness of the sequences.

I've only ever seen the proof in the general case, actually. The idea is to show that any two free (or projective) resolutions of $A$ are chain homotopic. Because chain homotopy is preserved by additive functors, the complex $\textrm{Hom}(P_\bullet, M)$ is well-defined up to chain homotopy, and any two chain homotopic complexes have the same homology.

Oy vey.

I guess it's a little bit easier for $\mathbb{Z}$ because $\mathbb{Z}$ is a PID or something...

1:18 AM

Are you feeling like helping me a bit with the details?

I can try.

Okay.

Suppose that $A,M$ are abelian groups, given $0\to P_1\stackrel{p}{\to} P_0\to A\to0$ a free resolution of $A$

Let us denote by $p_*$ the precomposition homomorphism from $Hom(P_0,M)$ to $Hom(P_1,M)$.

Then $Ext(A,M)=Hom(P_1,M)/im p_*$.

So far, so good?

Yes.

Now to show that this is independent of the choice of resolution suppose $0\to P_1\stackrel{p}{\to}P_0\to A\to0$ and $0\to F_1\stackrel{f}{\to}F_0\to A\to 0$ are two free resolutions of $A$, we want to show that the $Ext$ defined by both resolutions are equal.

To save the space let us denote the $Ext$ defined by the $P$'s as $Ep$ and the one defined by the $F$'s as $Ef$.

OK. So how do you proceed?

1:28 AM

Now, since $P_0$ is projective and $F_0$ is mapped onto $A$ there is $d_0\colon P_0\to F_0$ which closes the square created using $id\colon A\to A$ creates.

Now I showed that we can use $d_0$ to define $d_1\colon P_1\to F_1$ such that the entire diagram is commutative.

Yes. And you can do the same the other way. The only problem is that the composition either way need not be the identity.

(By the entire diagram I mean the two short sequences and the $d_1-d_0-id$ connecting them)

Correct.

But you can use a more sophisticated argument to show that they are quasi-inverses in the sense of chain homotopy.

Let's denote the maps $c_i\colon F_i\to P_i$ for those which we got.

What does quasi-inverse mean?

It means there is a chain homotopy to the identity.

1:33 AM

Ah.

I also know how to define $g\colon Ep\to Ef$ and $h$ in the other direction.

But I'm not sure how to show that either is an isomorphism.

Well, it's a general fact that if two maps are chain homotopic, then the induced maps in homology are equal.

I think that I even saw a proof of that at some point in one course or another... but my memory is not what it used to be.

Well, it's enough to prove it in the case that the two maps are null homotopic. It's an easy element chase.

Essentially that $h-g=0$?

Well, in your notation, $c \circ d - \textrm{id} \sim 0$ and $d \circ c - \textrm{id} \sim 0$.

It follows then that $g \circ h - \textrm{id} = 0$ and $h \circ g - \textrm{id} = 0$.

1:46 AM

Indeed it follows.

So now what I need to do?

Well, you showed that $Ep$ and $Ef$ are isomorphic. What more do you need to do?

No, I mean... that "easy element chase" you speak of...

I'm literally helpless right now... it's 4am and I've been up since yesterday morning.

I also spent the past few hours banging my head against these diagrams...

So it's all... murky.

Ah, the element chase. Well, let $B_\bullet$ and $C_\bullet$ be chain complexes, and let $f_\bullet : B_\bullet \to C_\bullet$ be a chain map homotopic to 0.

By homotopic to 0 you mean? I'm not sure I understand this concept outside topological spaces...

That means there is a map $s_n : B_n \to C_{n+1}$ such that $f_n = c_n \circ s_n + s_{n-1} \circ b_n$, where $b$ and $c$ are the differentials of $B_\bullet$ and $C_\bullet$, respectively.

1:55 AM

Okay...

We need to show that $H_n (f) = 0$.

So it would be the maps I pulled to the resolutions if I had closed the $A\to A$ using $0$ instead of $1_A$?

No, this is a separate lemma.

I swear... one day I will invent a time machine and kill all those jerks with their diagrams. I will kill them good, I tell ya!

It would be much more painful without diagrams, I assure you...

1:58 AM

Can you try and dumb this chase down to the case at hand? I get easily lost in all those indices.

No, I find special cases more confusing.

In set theory I would probably agree with you... :-P

We could also do the whole thing without indices. Let $B = \bigoplus_n B_n$ and such.

Nah.

So essentially, this all nightmarish hell I had to go through in AT has come back to bite me again?

Oy vey.

Homological algebra is homological algebra, whether you do it for AT or AC...

2:04 AM

Hell is not other people. Hell is homologies and cohomologies.

I still don't get why people keep insist that I will copy solutions from books, I mean the teachers.

It's like "We don't want to teach you this, but we want to feel that you've seen the proof. Write it from this book on this piece of paper."

In my experience, copying out proofs has a mysterious beneficial effect on understanding...

but perhaps that's because writing it out makes me stop to fill in missing details.

@ZhenLin Yeah, but it's a common knowledge that once I submit this (after copying) everything goes bye-bye in my memory.

Hell, I can hardly remember set theoretic proofs which I use on a non-regular basis, why would I make an effort to remember a proof that no one really wanted to teach me??

Indeed. Unfortunately for me, I have these nasty things called "exams" in which my memory is tested...

@ZhenLin This is why I consistently failed about half my exams during the undergrad :-D

Okay, no. This is not gonna work. I cannot see how the element chase with the chain complexes applies in this case.

Well, perhaps it doesn't help that $\textrm{Hom}(-, M)$ is turning our chain complexes into cochain complexes...

but you know, just reverse all the arrows. :p

2:16 AM

I have $c_i$ and $d_i$ and $c_i^\ast$ and $d_i^\ast$, and I have no idea what I am supposed to show anymore in this goddamn proof.

If you show that $c_i$ and $d_i$ are quasi-inverses, then so too are $c_i^*$ and $d_i^*$. Then by the dual of the lemma, these induce isomorphisms in cohomology.

The dual of the lemma?

The lemma which shows that chain homotopic maps induce the same maps in homology.

Another lemma, and another lemma... and another lemma... and then another lemma.

I'll just adjoin a copy of goddamn Rotman in the middle of my work, why not?!

These are all well-known, and you can surely omit their proofs (if not their statements too)!

2:21 AM

I have no idea.

The teacher was really messy in class and I have absolutely no idea what we proved or did not prove in class.

Supposedly we proved that Ext does not depend on the choice of resolution, I don't know... maybe we didn't.

He told me that in the $\mathbb Z$-modules case I don't need all that machinery and I can just do it by hand.

Hm, PlanetMath defines quasi inverse when $f\circ g\circ f=f$.

Hmmm... yes, I suppose so. You just expand all the lemmas in special cases and inline the proofs. :p

The word ‘quasi-inverse’ is used in many contexts.

@ZhenLin You know, I'm sure it's really funny to you but I'm going to smash the monitor and trash the office soon...

It's 4:30 and I've done practically nothing for the past 8 hours or so except writing the same diagram from different angles.

I wanted to finish this about six hours ago and go have a beer.

I don't know of any completely bare-hands proofs, and I've never seen one.

If I have that $\varepsilon\circ c_0\circ d_0=\varepsilon$... is that good?

Where $\varepsilon\colon P_0\to A$ is the surjective map at the end of the resolution.

2:39 AM

Hmmm... possibly. But that only shows that the $H^0$ of the two resolutions agree.

I also have that $f\circ d_1\circ c_1=d_0\circ c_0\circ f$.

Which makes no sense at all.

:|

That's the chain homotopy showing that $d \circ c \sim 0$, which is deeply troubling because that shouldn't be true.

So it makes sense after all!

Recall that $f\colon F_1\to F_0$.

It's not the homotopy between the chains, but rather the map inside the $F$-chain.

Actually, scratch that, it isn't relevant.

What isn't?

2:43 AM

What I said.

What about what I said?

It's just a formal consequence of composing two chain maps.

It gives the impression of being relevant to something!

Does it get me somewhere?

No, I doubt it.

Hm. Not sunrise yet, but the birds are waking up.

So let's go through that part again. I have $p\colon P_1\to P_0$ and $f\colon F_1\to F_0$, both injective. I also have $d_i\colon P_i\to F_i$ and $c_i\colon F_i\to P_i$ which commute with the $f$ and $p$ (and the surjections onto $A$).

Am I trying to show something on the $c_i$ and $d_i$ or on the $d_i^\ast$ and $c_i^\ast$?

2:47 AM

Hi, Asaf, having fun chasing diagrams? :p

No.

No progress for like 8 hours.

What do you try to do?

Well, the ultimate goal is to show that the induced maps in cohomology are isomorphisms. To do that the easiest way is to show that $c_\bullet$ and $d_\bullet$ are mutually quasi-inverse.

Show that Ext is independent of the choice of free resolution.

Mariano Suárez-Alvarez

you do know that two proj. resolutions are homotopic?

2:49 AM

@MarianoSuárezAlvarez I have no idea whether or I know that or not, or whether I should include that in my proof.

Mariano Suárez-Alvarez

you should

that's the key point :)

What exactly do you mean by homotopic?

Mariano Suárez-Alvarez

two maps of complexes $f, g:X\to Y$ are homotopic

I am going to use my last espresso capsule, be back in two minutes.

Mariano Suárez-Alvarez

this is surely defined in Rotman (or any other homological algebra textbook...)

and is not the kind of thing chat thingies like this work best for!

2:55 AM

THOSE $%&*($) A&*^HOLES!&*(@#$^%@#&*(^

Someone locked the kitchen!

Okay.

Rotman. Let's see...

What is a very simple program to write a pdf, or to even convert simple latex code from math.se to pdf ?

Is my attempted answer here correct: math.stackexchange.com/questions/124320/… ?

@tb Why are you up?

I am looking for some way to deliver my homework in latex, but I know no latex beyond posting on math.se

@AsafKaragila don't ask...

Mariano Suárez-Alvarez

3:02 AM

@Holowitz, latex :)

@tb Too late... :-P

Mariano Suárez-Alvarez

learn latex

and the use latex

Yeah, use a condom.

Mariano Suárez-Alvarez

the $\frac{\mathrm{usefulness}}{\mathrm{effort}}$ ratio of doing so is close to infinity.

3:20 AM

Hi guys

8k!

well congrats

Oh look, sunrise.

3:36 AM

@ZhenLin Congrats!

Thanks! But how did you get that picture? I'm at 8002 now...

@ZhenLin Not when I snapped the image

ah

I smell a downvote?

@AsafKaragila where?

3:40 AM

In those huge bags I have underneath my red and sleep deprived eyes...

Have a look at this

@RajeshD a typo: intezer

thanks @leo

@leo : BTW its not a typo, i should confess that its a spelling mistake on my part

3:57 AM

To allow an edit if you were downvoted?

Hmmm..LOL

AGH%$UIY

I hate the sparrows in the morning.

They are so friggin vocal, it drives me insane, especially after staying up all night.

I like their sounds

Sparrows? They are just annoying. I love blackbirds, though.

but not when deprived of sleep

4:01 AM

Blackbirds sing at dusk, though.

1 hour later…

5:02 AM

Hi @azarel

Ugh. This is madness, I feel that I've proved what I needed to prove, but I just don't know it yet.

5:23 AM

Hi, @AsafKaragila

How are you?

Not bad

How about you?

I think that I am finally catching up with this unholy diagram.

How I hate algebra, it's uncanny.

Kannappan Sampath

Hi @Asaf Algebra is fun, you know! Why do you think it is uncanny? I enjoy doing it!

Hi @azarel

Not the algebra is uncanny. The amount of hatred I feel towards it is uncanny...

Kannappan Sampath

5:27 AM

Oh I see.

Hi @KannappanSampath

@azarel Tell me something, my good man. Do you know Michael Hrusak?

yes I do

One of set logicians in my department once told me about him, he said that his thinking is jumpy in the sense that he can make seemingly unsubstantiated arguments which are true a posteriori.

And that he's really good too, which makes it even more magical.

I guess

5:32 AM

Well, I don't know him so I'm asking whether or not the description is somewhat accurate...

but that descriptions fits several people

Such as?

like justin moore, paul larson

are you in the same university as Shelah?

No, but I go there on a weekly basis.

I attend a course given by Magidor in descriptive set theory.

I'm in BGU right now, I do plan on doing my Ph.D. in Jerusalem though.

Nice

5:37 AM

Of course, first I need to catch this diagram... :-P

hahaha

No, seriously. This is the last thing I need to do before I can start working on my thesis... :-)

Hooray!

I managed to prove that $g$ is onto!

Kannappan Sampath

What to do with this?

Kannappan, it's time I tell you a little story about the birds and the bees.

Kannappan Sampath

@AsafKaragila Good! You see, you did not prove that for the past 8 hours because it was DIAGRAM CHASING. Now it is just diagram chasing and you're done!

@AsafKaragila I am willing to listen to it. : )

5:45 AM

Bees work very hard, but then come the bee eaters and eat the bees, those which survive get colony collapse syndrome and die. The birds are later electrocuted on high power lines. So the lesson here is that everybody dies, but animals only have to chase one another and not diagrams.

Kannappan Sampath

Amusing! : P

I am going to take a walk outside. The fresh air might do me some good, maybe I'll find an open food stand with actual food (most of them get the food only an hour from now...)

6:11 AM

Morning.

Not a good one though.

Whose idiotic idea was it to switch clocks twice a year? (<-resisting the temptation to put more than one ?)

Now time for the most important meal of the day: the first cup of coffee.

7:07 AM

Plane $T$ nears the surface $S: \int_0^1 \frac{e^{xzt}}{x-y+z+t} dt =\ln(2)$ in a point where $z\gt 0$. What is $T$?

What is the point with this q?

(Source problem 3 here.)

7:22 AM

So $\nabla S \times T =0$ at some point $t_{0}$?!

$\partial_x \left( \int_0 ^1 \frac{e^{xzt}}{x-z+z+t} dt \right)=?$

Moved to q here.

8:04 AM

I get exponential integral there, more here. Shortly $\int e^t ln(t) dt=?$

8:44 AM

@rob: are you here?

$$
\begin{array}{ccc}
\mathscr S & \models & \Psi
\\
\downarrow & \uparrow &\downarrow
\\
\tilde{\mathscr S} & \models & \tilde{\Psi}
\end{array}
$$

9:19 AM

@Jonas: hi

@Ilya Hi.

@Jonas: are you experienced with LaTeX beamer?

A bit, yes. But I need to go now :-).

@JonasTeuwen then see you later

Hi folks, good morning/afternoon whichever applicable

Hey @Ilya

9:26 AM

@Rajesh: hi

Whatz up ? How was the weekend

I've got a flue

but the weather is nice

sorry, I'm a bit busy with a presentation

Ilya i had a watman encounter

Yay hello teddy

ok....best of luck

Hi all

9:33 AM

Hi

Heh must ve been a figment of the imagination of my phone

what do you mean?

No you are here. Thought youd disappeared

I was here briefly about ten minutes ago.

How's your head? I saw that you were up all night

9:41 AM

Hi @tb

@Matt, @tb: hi

Hi Ilya, Rajesh

Hi ilya

hm... matt

@tb: what are your favourite animals? not necessary pets

@MattN I think that's the best I can do at the moment for answering your first question.

9:47 AM

I cant give you a carrot for that. Its NARA : )

@Ilya Hi.

And @tb @MattN Hi.

@JonasTeuwen em, hi again :) and what are you favorite animals?

Hi jonas

@Ilya Cats and catlike thingies. Monkey's are also cool.

@JonasTeuwen cool, I also love cats (except lions, though). This Saturday we were to the Rotterdam zoo - and I was sad since the tiger was hiding in the bushes

also the leopard was quite inactive and manul didn't appear at all

9:52 AM

@MattN I know very well. :) I'm not in the mood of elaborating too much, but this sums it up quite well :/

And im sad that there are people who financially support animal torture by going to the zoo

@MattN I'm not happy to see them in cages

Hmm, that's a bit controversial. Tigers in the wild are almost a goner for example. But oh well.

@Matt, and talking about a bit implicit things - maybe we currently do support the starving in Africa: talking in the chat (instead of going there and saving their lifes), receiving too high salary etc. And I'm not kidding

@RajeshD I've seen it.

9:56 AM

@JonasTeuwen why is tiger a goner in the wild?

@tb thanks for notifying....waiting for your reply

@Ilya Because they kill it :(.

@JonasTeuwen not only tigers

Phones gonna die soon

@tb : i forgot to mention, it does not take any of your time for thinking...its not mathematical question, just plain suggestion

10:01 AM

Oh lord, save me from this diagram! >:(

strange talk here today

Hm, no answer. I guess no one is there...

@Asaf: where?

I feel like Antonius Block.

@AsafKaragila what diagram?

10:02 AM

@tb : if you are still not able to spend a little time...its perfectly ok

@AsafKaragila that's a nice movie, although the coda is a bit surprising

@tb Same one... :|

@Ilya Yes, one of my favorites.

@AsafKaragila independence of Ext of resolutions?

@tb T_T

I take that as a yes :)

10:04 AM

It is no longer the independence of Ext. It is the independence of me from non set-theory stuff for the rest of the degree (modulo teaching calculus...)

I'm not aware of such a generalization of that result :)

It follows directly from that result...

So, instead of whining: a chain map between chain complexes gives a map between their cohomologies, right?

I don't know.

@Ilya I know.

10:14 AM

@AsafKaragila That would be the first thing to prove.

Just write down the definition of homology as cycles modulo boundaries.

Oh, apparently I have that already.

Next you need to think about why chain homotopic maps give rise to the same map in homology.

$f - g = dh + hd$ and a map of the form $dh + hd$ induces zero in homology.

Finally, the uniqueness of resolutions up to chain homotopy equivalence tells you that the homology of a resolution doesn't depend on the resolution.

..but only on the object you resolve.

I take it that $d$ is the boundary map and $h$ is the homotopy?

Exactly.

This is before or after moving to Hom(P_i,M) sequences?

10:21 AM

before.

So $f$ and $g$ are?

Just take any two chain complexes, $C$ and $D$ and chain maps $f,g: C \to D$.

Oh.. Oh.

This means $d_D f = f d_C$ and similarly for $g$.

In my case I have $f=id$ and $g$ is some composition I extracted from the projectivity of the groups in the resolution.

10:23 AM

Yes.

But wait with that for a moment.

So: again: is it clear to you why a chain map $f: C \to D$ gives you a map $f_\ast: H(C) \to H(D)$?

If $z$ is an $n$-cycle in $C$ then $d_n z = 0$ by definition. The relation $fd =df$ tells you that $f(z)$ is an $n$-cycle in $D$ because $df(z) = f(dz) = 0$.

Yes, that I know. I still have nightmares from AT...

Now if $b$ is a boundary, it is of the form $dc$ for some $c$. Thus $f(b) = f(dc) = df(c)$, so $f(b)$ is a boundary, too. Et cetera.

Are we still pre-$Hom(-,M)$?

someone have a look at this please : I do not understand what's wrong with it ?

@AsafKaragila yes. Still with chain maps $f,g: C \to D$ and what they do on the level of homology.

10:31 AM

@tb Okay, yeah, I think I understood what you mean. Yes, I did that already... several times... :-)

Now suppose you have that $f$ and $g$ are chain homotopic chain maps $C \to D$. Then $f-g = dh - hd$ for some maps $h$ going "diagonally". If $z$ is a cycle in $C$ then $(f-g)(z) = (dh + hd)(z) = dh(z) - hd(z) = dh(z)$ because $dz = 0$. So cycles are mapped to boundaries by $f-g$, hence they are mapped to zero in homology.

This means that $f-g$ is zero in cohomology, or in other words, $f$ and $g$ give rise to the same map $f_\ast = g_\ast : H(C) \to H(D)$.

@Ilya I am now :-)

@Jonas: congrats for the bounty!

So now removing all the abstraction from the problem.

I really just to solve this one particular case and that's it.

@AsafKaragila Now after proving that projective resolutions are unique up to chain homotopy equivalence, we are in position to get where you want.

10:39 AM

Well. Can't I just inline that lemma?

I'm not sure I can contain another lemma, mentally at least.

Namely: $\operatorname{Hom}{({-},M)}$ sends chain complexes to cochain-complexes, chain maps to cochain maps and chain homotopic maps to chain homotopic maps.

This means that the homology complex $\operatorname{Hom}{(P_\bullet,M))}$ does not depend on the chosen resolution $P_\bullet$.

@AsafKaragila Of course, you can do this in a down to earth fashion (i.e. completely explicitly), but it doesn't really get any easier. You'll have to show the same things anyway.

I don't care it doesn't get easier.

I really just want to take those diagrams I defined with all those homomorphisms going from one place to another and prove these things directly on those.

You are only interested in $\operatorname{Ext}^1$, right?

Yes.

Let me bring you up to speed.

I have $A$ and both $0\to P_1\to P_0\to A\to 0$ and $0\to F_1\to F_0\to A\to 0$ are free resolutions of $A$.

The maps are $p$ and $f$, since we only really care about the map from $P_1\to P_0$ and similarly in the $F$'s case no need to number those, we can name the maps onto $A$ as $p_0$ and $f_0$ for consistency sake.

Using projectivity of $P_0$ and that $f_0$ is onto $A$ we define $d_0\colon P_0\to F_0$, and then it is easy to show there is a well defined map which closes the diagram commutatively, $d_1\colon P_1\to F_1$.

So you lift $p_0: P_0 \to A$ to any old map $g_0: P_0 \to F_0$, that is $p_0 = f_0 g_0$.

10:47 AM

Similarly, define $c_i\colon F_i\to P_i$.

($d$ is a very bad name, but okay, that's right).

Okay.

We denote by $\bullet^\ast$ the induced maps after applying $Hom(-,M)$ to all that diagram.

Okay, so you need to check that the maps $\bullet^\ast$ (or rather what they induce in homology). Don't depend on any of the choices.

Now we define $Ef$ and $Ep$ the $Ext$ which would be defined by each resolution, namely $Hom(F_1,M)/Im(f^\ast)$ and similarly for $Ep$.

Right.

10:50 AM

@tb What choices? I take their well-definability for granted... $Hom$ is a functor, no?

@tb I also defined the obvious maps from $Ef$ to $Ep$ and vice versa, now I have the entire three-parts diagram is commutative and exact at $Ef$.

@AsafKaragila It is. But you need to check that if you choose other maps $d_0'$ and $d_1'$, etc. that doesn't really matter.

@tb That's the part where we check that homotopic maps induce the same maps?

Exactly.

So suppose you're given two choices $d_0, d_1$ and $d_0', d_1'$.

Now I have the map from $Ef$ to $Ep$ and I want to show it is isomorphism.

Wait a second, that's automatic after what I'm going to say.

Then $f_0 (d_0 - d_0') = 0$, so the image of $d_0 - d_0'$ lies in the image of $f_1$ by exactness of $F_1 \to F_0 \to A$.

This tells you that there is a map $h_0: P_0 \to F_1$ such that $f_1 h_0 = d_0 - d_0'$.

10:58 AM

@tb I've got a bounty? 8-).

@JonasTeuwen The $W^{1,1}$-thingie.

I've got a bounty but not an upvote. Hmm.

Okay, nice.

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