Room for AKE and Limitless

Limitless

1:23 AM

Ok. Thanks for your time. What's he getting at?

AKE

I don't have Basic Algebra -- I've waded through some of his others (not terribly enjoyed them either).

However, why do you think it is the "other interpretation" -- and which one do you mean?

Limitless

I believe that he is asking: What are the consequences if $\alpha\beta=1_S$, $\beta\alpha=1_T$, and $\beta$ is unique?

AKE

Thinking...

Limitless

I think this because this makes sense as a question as it directly has an answer: $\alpha$ is bijective via the definition of a bijection.

AKE

Yes but that would be trivial, wouldn't it?

Limitless

1:26 AM

I feel he simply "constructed" the definition of bijection within the question and is asking the reader if he or she understands.

Perhaps.

AKE

Does he have any similarly "trivial" exercises in that book?

Limitless

But it is the second exercise of the introduction chapter.

Yes.

Well

He has one similar to this that comes to mind. It is a bit more in depth, but it reads just like this: He constructs a definition via its elementary parts.

I will get the link to my math.SE question regarding that one.

AKE

No that's ok.

So it may be you're right, however, when it comes to N Jacobsen, I prefer to think that he's trying to get at something else...

So let's try the other way...

Limitless

What happens then? :-)

AKE

Looking back at your notation... hang on.

Limitless

1:30 AM

My thoughts would be that there is some sort of way of showing that if $\beta$ is unique then $\alpha$ is unique and it is the only such bijective map that satisfies $\alpha\beta=1_T$ in one case and satisfies $\beta\alpha=1_S$ in another case.

However, this doesn't really make sense to me.

AKE

Hm, I was thinking along the lines, what if a mapping between two sets is injective (one-to-one), and we have that the partial inverse map $beta$ is unique...

going further, what is the uniqueness giving us -- what would happen if we relax the uniqueness constraint... can we construct a mapping such that there are two $beta$'s?

Limitless

The issue is that we need injective and surjective to be simultaneously true in order to have a bijection. The premise of transforming "$\beta$ is unique and $\alpha\beta=1_T$" into "$\alpha$ is injectve" and "$\beta$ is unique and $\beta\alpha=1_S$" into "$\alpha$ is surjective" confuses me.

Do you see what I mean? I am deconstructing this to the very basics: propositions. And I can't seem to link the two propositions.

AKE

Deconstruction to the two propositions is exactly what I was getting at in my answer -- that is the explanation of how I am parsing it.

Limitless

My issue is

AKE

Now, as to how we show these -- well, that, in my opinion, is what the exercise is about.

Limitless

1:36 AM

In this deconstruction, we may have created an exercise which isn't there. :-P

AKE

So, we have the choice between a non-exercise and an uncertain enterprise :)

Limitless

I honestly do not know how to answer this exercise if that is how the question is to be interpreted.

Hah.

Nice point.

We either know or we know nothing. I'm LOLing.

AKE

If this is homework and you're pressed for time, I'd go with the non-exercise interpretation, and just do the usual explaining that you understand the question...

If, however, you're doing this for your own enjoyment...

Limitless

I'm doing this to learn.

AKE

then it actually becomes (in my opinion) one of those side tangents that you chew on while you continue onwards...

And the chewing consists of the following:

either showing that aha! the two apparently unrelated concepts are in fact related (i.e. uniqueness + injective => surjective and therefore bijective, and uniqueness + surjective => injective and therefore bijective)...

or constructing a counter-example!

Limitless

1:40 AM

Hm . . .

I'm thinking of a counter example.

AKE

One of the first things I try when faced with frustrating ambiguities in abstract math books (and there are more than one wishes!) is to try to come up with a counter-example...

Usually the effort of doing so becomes quite rewarding as I learn that bit of the terrain really well. Sometimes it goes somewhere, sometimes not, but usually one reaches a kernel on which the entire thing turns. And at that point, you now have a more specific question to either focus on or re-post on SE :)

That's not to dodge the challenge -- but that's the reason why I parsed the question the way I did...

Limitless

Let $S=\{s_1,s_2,s_3\}$. Let $T=\{t_1,t_2,t_3,t_4\}$. Define $\alpha: S \to T$ such that $s_i \mapsto t_i$. The only $\beta$ such that $\beta\alpha=1_S$ is $\beta: T\to S$ such that $t_i \mapsto s_i$. $\alpha$ is still not bijective.

How's that?

I wrote it off my thoughts without referencing notation. Let me make sure it's what I meant.

AKE

Yep, that's what I had scratched out as well. Let's consider for a moment...

Limitless

Looking at my notations, that seems solid. Do go on.

AKE

Ok, how about this one (dropping the subscript notation entirely):

Limitless

1:46 AM

(We can honestly reduce it to simply $S=\{s_1\}$ and $T=\{t_1,t_2\}$).

Yes

I see your point. That's even more elegant.

$S=\{s\}$, $T=\{a,b\}$.

AKE

S: 1 2 3
T: A B C D
Map: 1 -> A, 2 -> B, 3->C
But now inverse map can be:
A->1, B->2, C->3, D-> ANYTHING

Limitless

$\alpha:S \to T$ with $s\mapsto a$ and $\beta: T\to S$ with $a \mapsto s$. Presto-bingo.

interesting, let me look

Your statement about the inverse map assumes that the inverse map acts on all elements of $T$. I do not think this is the case.

We are getting to some very rough, rough fundamentals.

Give me a second and I will explain.

AKE

So alpha map is injective (one-to-one), but there can be many beta maps that pull-back images under alpha to themselves.

All that we have in the problem is that beta map is defined as from T to S. There is no constraint that says that T and S are in any way related or have the "same number of elements", only that beta takes elements of T (assume the entire T) and maps them to S.

Limitless

See--that is the issue, I think. You are assuming the entire $T$.

AKE

The inverse mapping above (with D -> anything) fits this, and satisfies beta*alpha = 1_s

Limitless

1:50 AM

Let me explain why that doesn't make sense to me.

AKE

Go on.

Limitless

Ok?

Alright:

All we know is that there is a map from S to T; we call this $\alpha$. Likewise, all we know is that there is a map from T to S; we call this $\beta$. When we presuppose $\beta\alpha=1_S$, here's what we have: An $s$ maps, via $\alpha$, to some $t$ for all $s$ in $S$. These $t$s, __which are not necessarily are of $T$__ since we do not know if $|S|=|T|$, then map back to the original $s$ via $\beta$.

From this, we can conclude that my counter example holds perfectly for the _general_ notion that it is trying to disprove.

AKE

Am ok up to the last sentence of first para...

Limitless

In the case that $|S|=|T|$, then $\alpha$ is a bijection. However, we don't know that.

AKE

But wait there: are you missing a key word in the underlined phrase?

Limitless

1:56 AM

My bad! I meant to say

AKE

(BTW -- let's agree to drop the latex notation -- it's just adding noise to the paragraph)

We're among friends here ;)

Limitless

"which are not necessarily all of T"

The issue is this: We can't say |S|=|T|.

AKE

Ok, good. Agreed with the whole paragraph. So then what?

Limitless

So. Hm.

Let's see . . .

AKE

From this we can conclude... please draw the conclusion without referring to anything prior...

in the chat I mean...

Limitless

1:58 AM

I'd say that $\alpha$ is not bijective unless $|S|=|T|$ given the original parameters.

oops, latex'd by reflex. Sorry, haha.

AKE

:)

Limitless

So, we have a nice conclusion here.

AKE

ok thinking...

Limitless

But, hm.

Something is twinging me. If we say some map, say, z, maps from L to M . . .

then does this mean that there exists a z(l) for all l in L?

In ALL cases of z

AKE

... right. Exactly. Which is the same as to say that beta is not unique unless |S| = |T|, because if T were bigger than S, then you could have this hang-about d element being mapped anywhere...

good god man! No more letters! We already have some good ones (alpha, beta, S, T, 1,2,3, A,B,C)

:)

Limitless

2:01 AM

hahahah

I think we've pieced it together.

We have to combine the very definition of a map.

You are right.

AKE

Yep.

Limitless

Jacobson is getting at something very deep here.

We have that, like in your example, D has to map to SOMETHING

AKE

Precisely.

Limitless

and the only case where this something is a specific element and not just 'something'

is when |S|=|T|

Christ.

AKE

not sure I follow the last sentence of somethings...

Limitless

2:02 AM

Yeah, it was terrible phrasing.

Let me rephrase

The only case where D maps to a specific element and not just any element is when |S|=|T|.

This is what it means to say beta is unique.

AKE

Yes -- I think of D as a free variable.

Limitless

I am going to post an answer

AKE

The condition of equal cardinality means that injection = surjection (and therefore bijection). But Jacobson has thrown something else quite cool into the mix -- the concept of uniqueness.

Limitless

Summarizing all of this and writing it in my lovely language

Do you mind?

Jacobson is essentially telling us to see that uniqueness implies equal cardinality which implies injection=surjection which implies bijection.

AKE

Go for it.

Though to be fair, have a look and if you see that the sentence is parsed the way I said, would ask that you +1 my answer ;)

Limitless

2:06 AM

Oh, that's completely fair

AKE

Well, nice chatting. Enjoy the rest of Jacobsen.