injective and surjective

From: Henno Brandsma Date: September 8, 2010 Subject: Re: Re: Re: injectivity and cancellation property.

In reply to "Re: Re: injectivity and cancellation property.", posted by BD on September 7, 2010:> Then does this also true that surjective function does not have left cancellation >property? but how?

A surjective and non-injective function has the right cancellation property, but does not have the left cancellation property.This is clear because we have shown that f 1-1 <==> f has left cancellation property.[in the category of sets!]

Henno

>Please thank you so much for your help>>>>In reply to "Re: injectivity and cancellation property.", posted by Henno Brandsma on September 7, 2010:>>In reply to "injectivity", posted by BD on September 7, 2010:>>>how to pv: if f:A to B has left cancellation property then f is injective>>>(I found the converse).Also can we find a counter example that f is >>> injective but has no right-cancellation property?>>>thanks in advance>>>>Suppose f:A -> B has the left cancellation property, so>>>>(*) for all maps g,h: C --> A, if f o g = f o h, then g = h.>>>>To see that f is injective: suppose not and let a != a' be 2 points in A, such that f(a) = f(a').>>>>Now define C = {0,1}, and g: C --> A by g(0) = a, g(1) = a'>>and h: C --> A by h(0) = h(1) = a.>>Then f o g = f o h, as both are constant with value f(a) = f(a').>>But g != h, as witnessed by g(1) = a' != a = h(1).>>>>We can also show that the right cancellation property is equivalent to surjectivity, so>>if we have an injective but not surjective map, say >>f: {0}--> {0,1}, defined by f(0) = 0, this will not have the right cancellation property.>>(use g,h: {0,1} --> {0,1} defined by g = id and h == 0: h o f = g o f but g != h)>>>>Henno

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Adam - 08 Jul 2004 14:49 GMTHi,

I am studying set theory on my own as a hobby. The primary text that I useis Proofs and Fundamentals: A first course in abstract mathematics by Bloch.Below is a theorem that I wish others to read and then provide constructivecomments about. It is a simple theorem, but I am just learning set theory,and so it is a good theorem for me to prove.

Theorem 4.4.4. Let A and B be non-empty sets, and let f: A -> B be afunction.(ii) The function f is injective iff f o g = f o h implies g = h for allfunctions g, h: Y -> A, for all sets Y.

Proof. Suppose f is injective. We assume that f o g = f o h. Since f was

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assumed injective, there exists a left inverse for f, which we denote by j:B -> A, such that j o f = 1_A. Applying j to both sides of our equation weget j o (f o g) = j o (f o h), and so (j o f) o g = (j o f) o h. Since j o f= 1_A, we get 1_A o g = 1_A o h. Thus, by the identity law, g = h, asdesired.Now suppose f is not injective. There then exists some b in B such therefor some x != y in A, f(x) = f(y) = b. Let Y = {1,2}. We define g asfollows: let g(1) = x, and g(2) = y. We define h as follows: let h(1) = y,and h(2) = x. Thus f o g = f o h, but g != h. The desired result followsfrom the contrapositive. QED.

Thank you, Adam.

> I am studying set theory on my own as a hobby. The primary text that I use> is Proofs and Fundamentals: A first course in abstract mathematics by Bloch.[quoted text clipped - 6 lines]> (ii) The function f is injective iff f o g = f o h implies g = h for all> functions g, h: Y -> A, for all sets Y.

If f injection, then from fg = fhf(g(x)) = fg(x) = fh(x) = f(h(x))g(x) = h(x) for all x; g = hby definition of injection f(x) = f(y) ==> x = y

Slightly less overhead than yours.

--ConverselyAssume f(x) = f(y).let g(Y) = {x}, h(Y) = {y}then fg(z) = f(g(z)) = g(x) = f(y) = f(g(z)) = fg(z) for all z in Yfg = fh; g = h; x = g(z) = h(z) = y; x = y QED.

About the same as yours but direct constructive proof withoutcontraposition. Notice Y can be any set /= nulset, even Y = {1}.So g(1) = x, h(1) = y will suffice which is a bit easier on the headthan the twist you did with Y = {1,2}

> Proof. Suppose f is injective. We assume that f o g = f o h. Since f was> assumed injective, there exists a left inverse for f, which we denote by j:[quoted text clipped - 7 lines]> and h(2) = x. Thus f o g = f o h, but g != h. The desired result follows> from the contrapositive. QED.

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David C. Ullrich - 08 Jul 2004 16:40 GMT>> I am studying set theory on my own as a hobby. The primary text that I use>> is Proofs and Fundamentals: A first course in abstract mathematics by Bloch.[quoted text clipped - 13 lines]

>>Slightly less overhead than yours.

But no sentences. Sentences good. Non-sentences bad.

>-->Conversely>Assume f(x) = f(y).>let g(Y) = {x}, h(Y) = {y}>then fg(z) = f(g(z)) = g(x) = f(y) = f(g(z)) = fg(z) for all z in Y>fg = fh; g = h; x = g(z) = h(z) = y; x = y QED.

Seriously, if I saw these two solutions on a homeword sethe'd get a perfect score, although maybe I'd wish he'dwritten a _little_ less. You'd lose a certain amountfor not writing in complete sentences and _also_ fornot clarifying what you were assuming at various points,for example "Conversely" should be "Conversely, assumethat fog = foh implies g = h."

On the other hand, I didn't notice reading his thatone could use a much simpler g and h. Probablywhat I'd be thrilled to see would be something likethis:

Proof. Suppose f is injective and f o g = f o h. Since f is injective it has a left inverse, which we denote by j. Now

g = (jof)og = jo(fog) = jo(foh) = (jof)oh = h.

[or maybe just "g = jofog jofoh = h" there.]

Now suppose f is not injective; suppose that x != y butf(x) = f(y). Let Y = {1}, and define g(1) = x, h(1) = y.Then fog = foh although g != h. QED.

My _guess_ is that writing out your "direct" versionof the converse, this completely, will make it longer.Let's see:

Conversely, assume that fog = foh implies g = h, andsuppose f(x) = f(y). Let Y = {1}, and define g(1) = x, h(1) = y. Then fog = foh, hence g = h, so x = y. QED.

No, it was about the same.

************************

David C. Ullrich


William Elliot - 09 Jul 2004 14:48 GMT> >If f injection, then from fg = fh> >f(g(x)) = fg(x) = fh(x) = f(h(x))> >g(x) = h(x) for all x; g = h> >by definition of injection f(x) = f(y) ==> x = y>> But no sentences. Sentences good. Non-sentences bad.

Outlines can use phrases and I make no claim other than presenting anoutline for a proof. However you may not fault me for not using easilyread printed text book style with spaces on both sides of = sign and oftenon both sides of + sign. Also like a text book I try to present formulasand mathematical expressions without the interruption of a new line.

> >Conversely> >Assume f(x) = f(y).[quoted text clipped - 4 lines]> On the other hand, I didn't notice reading his that> one could use a much simpler g and h.

Yes, if anything, within my capacity I would be a better proof checker(including typos) than a proof writer.

> Probably what I'd be thrilled to see would be something like

> Proof. Suppose f is injective and f o g = f o h. Since f is> injective it has a left inverse, which we denote by j. Now[quoted text clipped - 3 lines]> f(x) = f(y). Let Y = {1}, and define g(1) = x, h(1) = y.> Then fog = foh although g != h. QED.

Indeed, your proof has the polished quality of an experienced authorwriting for a text book.

> My _guess_ is that writing out your "direct" version> of the converse, this completely, will make it longer.[quoted text clipped - 4 lines]>> No, it was about the same.

As little is to be gained by indirect proof, the direct is preferable forbeing constructive.


Adam - 09 Jul 2004 20:36 GMT> Seriously, if I saw these two solutions on a homeword set> he'd get a perfect score, although maybe I'd wish he'd[quoted text clipped - 3 lines]

> for example "Conversely" should be "Conversely, assume> that fog = foh implies g = h."

I'm glad that my proof would get a perfect score! I've come a long wayfrom not even knowing logic or what a set was.

> Proof. Suppose f is injective and f o g = f o h. Since f is> injective it has a left inverse, which we denote by j. Now>> g = (jof)og = jo(fog) = jo(foh) = (jof)oh = h.>> [or maybe just "g = jofog jofoh = h" there.]

At least to me, that is the exact some proof of mine, but without theextra clarification. I wanted to include each step just to be overlyprecise.

> Now suppose f is not injective; suppose that x != y but> f(x) = f(y). Let Y = {1}, and define g(1) = x, h(1) = y.> Then fog = foh although g != h. QED.

This proof is much nicer to read than my own.

The x and y that you use are not properly defined, at least how I havelearned to write proofs. You should write something like "Let x, y in A."Also, there should be some justification for assuming that two such distinctelements exist in the set A; this is because f is not injective. I assumeyou just didn't write these parts because of your advanced knowledge, but Itry to include any and all assumptions in my proofs. I sometimes forget, butthen my proofs are not correct to me. Even if they are obviously true orpeople understand what I mean. Of course, I could be wrong about mysuggestions for your proof, but I'm just being honest in how I see things,and not meaning to nitpick. I trust that if I am incorrect, you will let meknow, so that I can correct a misunderstanding.

> Conversely, assume that fog = foh implies g = h, and> suppose f(x) = f(y). Let Y = {1}, and define g(1) = x,> h(1) = y. Then fog = foh, hence g = h, so x = y. QED.

Again, I see a need to specify what x and y are. It's plainly obviouswhat you mean, but I think it should be written to present all assumptions.

Thank you very much, Adam.


Arturo Magidin - 09 Jul 2004 20:43 GMT>> Conversely, assume that fog = foh implies g = h, and>> suppose f(x) = f(y). Let Y = {1}, and define g(1) = x,>> h(1) = y. Then fog = foh, hence g = h, so x = y. QED.>

> Again, I see a need to specify what x and y are. It's plainly obvious>what you mean, but I think it should be written to present all assumptions.

As theorems get more complicated, trying to write everything downbecomes not only more cumbersome, but very lengthy, complicated, andsometimes even obfuscating to the main point. Think about, forexample, all the steps you skip in algebraic manipulations as a matterof course! If you have

5 = x + 2

you conclude that x = 3. But if you wanted to write every step,you would need to do something like:

Since 5 = x + 2, we can add -2 to both sides:

5 + (- 2) = (x+2) + (- 2).

Since addition is associative, we can rewrite the right hand side as

(x+2) + (-2) = x+ (2+ (-2)).

Since 2+ (-2) = 2 - 2 = 0, we have that the right hand side is

(x+2) + (-2) = x+ ( 2 + (-2) ) = x+ 0.

But 0 is the additive identity of the sum, so x+0 = x. Therefore,using the transitivity of the equality, we have:

5+ (-2) = x.

And 5+(-2) = 5-2 = 3. So the left hand side is equal to 3. By thetransitivity of equality, we have:

3 = x.

And I skipped some steps along the way, too.

If you don't feel comfortable skipping the steps, by all means, donot! The rule of thumb for skipping steps:

(1) You are comfortable doing so;(2) The remaining text is clear or obvious to you;

and most importantly:

(3) You are both able and willing to provide those steps if asked todo so.

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======================================================================"It's not denial. I'm just very selective aboutwhat I accept as reality."--- Calvin ("Calvin and Hobbes")======================================================================

Arturo [email protected]


Adam - 09 Jul 2004 21:00 GMT> As theorems get more complicated, trying to write everything down> becomes not only more cumbersome, but very lengthy, complicated, and> sometimes even obfuscating to the main point. Think about, for> example, all the steps you skip in algebraic manipulations as a matter> of course! If you have

Hi,

I do understand about not including every logical step in a proofbecause some are obvious to any mathematician at a certain level. As peoplehave written before, mathematicians wouldn't even write a proof for thisbecause it would be easy for them to prove themselves. The author of my bookwrote that proofs should be written with the intended audience in mind andso suited to their knowledge and abilities. For instance, he writes that ajustification for the quadratic equation would not required when writing aproof for people of a certain knowledge because such information would becommonly known or easily looked-up in a textual resource. I subscribe tothat view as well.

When I present my proofs to this news group, I fully understand that Ihave less mathematical knowledge than most readers, and am more thangrateful for others to take the time to help me, as I know my questions mayseem rather silly and trivial at times. However, I can not write my proofsto suit the general knowledge of this news group because I, myself, do notknow any more mathematics that what I write in my proofs. Many times myproofs have been dramatically simplified by others since they can use theirknowledge and experience to great affect. When that is done, as was the casefor Theorem 4.4.4 (ii), I learn a lot and am all the wiser for it. Forinstance, I thought the simplist way to prove part (ii), when f is notinjective, was to have a Y with two elements, however others showed how a Ywith a single element could be used. Obviously I had the write idea, whichis a good thing, but not the best idea. I'm working to further my

understanding of mathematics and not just my ability to write proofs;however, proof-writing is also important to me.

Again, thank you for your help, Adam.


David C. Ullrich - 10 Jul 2004 00:21 GMT>> Seriously, if I saw these two solutions on a homeword set>> he'd get a perfect score, although maybe I'd wish he'd[quoted text clipped - 17 lines]>extra clarification. I wanted to include each step just to be overly>precise.

Fine. No problem with that at all. The steps I left out werethings that it seemed to me could not possibly be misunderstood:

>> Now suppose f is not injective; suppose that x != y but>> f(x) = f(y). Let Y = {1}, and define g(1) = x, h(1) = y.[quoted text clipped - 4 lines]> The x and y that you use are not properly defined, at least how I have>learned to write proofs. You should write something like "Let x, y in A."

Officially, yes. But in the very same sentence I mention f(x) and f(y); it this point if the reader has any competence at all heknows I meant for x and y to be elements of A, because otherwise"f(x)" and "f(y)" make no sense.

>Also, there should be some justification for assuming that two such distinct>elements exist in the set A; this is because f is not injective.

Huh? Here you're not reading carefully, or something: I _said_ "Assumef is not injective" right there in black and white.

> I assume>you just didn't write these parts because of your advanced knowledge, but I[quoted text clipped - 4 lines]>and not meaning to nitpick. I trust that if I am incorrect, you will let me>know, so that I can correct a misunderstanding.

When you're starting out it's better to include too muchthan too little. When you get to be a little more experienced,so you and the reader can fill in the obvious details, includingtoo much makes things harder to read.

>> Conversely, assume that fog = foh implies g = h, and>> suppose f(x) = f(y). Let Y = {1}, and define g(1) = x,>> h(1) = y. Then fog = foh, hence g = h, so x = y. QED.>> Again, I see a need to specify what x and y are.

This x and y are the same as in the previous sentence! Ifthey'd changed without notice that would be a bad thing.

>It's plainly obvious>what you mean, but I think it should be written to present all assumptions.>> Thank you very much, Adam.

************************

David C. Ullrich


Marc Olschok - 08 Jul 2004 15:58 GMT> Hi,> [quoted text clipped - 8 lines]> (ii) The function f is injective iff f o g = f o h implies g = h for all> functions g, h: Y -> A, for all sets Y.

At least for statement (ii) , the restriction "A and B non-empty"is not needed. Was it used somewhere else in the Theorem?

> Proof. Suppose f is injective. We assume that f o g = f o h. Since f was> assumed injective, there exists a left inverse for f, which we denote by j:> B -> A, such that j o f = 1_A. Applying j to both sides of our equation we> get j o (f o g) = j o (f o h), and so (j o f) o g = (j o f) o h. Since j o f> = 1_A, we get 1_A o g = 1_A o h. Thus, by the identity law, g = h, as> desired.

How exactly, did the text define "injective function" ?The property you used (existence of a left inverse) is equivalentto "injective" in case of a nonempty A, but if the text defined"injective" via "for all x,y in A : f(x)=f(y) ==> x=y",then a proof could proceed from that definition by proving thecontrapositive along the line"If g not equal h, then there is a c in Y with g(c) not equal h(c) and ..."

This has the advantage, that such a proof would also work in situationswhere "f injective" ==> "f has a left inverse" is not applicable.

> Now suppose f is not injective. There then exists some b in B such there> for some x != y in A, f(x) = f(y) = b. Let Y = {1,2}. We define g as> follows: let g(1) = x, and g(2) = y. We define h as follows: let h(1) = y,> and h(2) = x. Thus f o g = f o h, but g != h. The desired result follows> from the contrapositive. QED.

The smaller Y={1}, with g(1)=x and h(1)=y already does it.

Altogether, the proof works (unless I overlooked something).

Marc


Arturo Magidin - 08 Jul 2004 18:55 GMT>> Hi,>> [quoted text clipped - 29 lines]>This has the advantage, that such a proof would also work in situations>where "f injective" ==> "f has a left inverse" is not applicable.

There is only one such situation, however: for f to be injective butnot have a left inverse requires that the domain of f be empty, butthe codomain nonempty.

But if we have that the domain of f is empty, and in addition we havethat f o g = f o h for some functions g and h, then the codomain of gand h must be empty, which means the domain of both is empty, and gand h are both the empty function from the empty set to itself.

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Marc Olschok - 09 Jul 2004 18:12 GMT>[...]>>> Theorem 4.4.4. Let A and B be non-empty sets, and let f: A -> B be a>>> function.>>> (ii) The function f is injective iff f o g = f o h implies g = h for all>>> functions g, h: Y -> A, for all sets Y.

>[...]>>> Proof. Suppose f is injective. We assume that f o g = f o h. Since f was[quoted text clipped - 3 lines]>>> = 1_A, we get 1_A o g = 1_A o h. Thus, by the identity law, g = h, as>>> desired.

>>[...] then a proof could proceed from that definition by proving the>>contrapositive along the line[quoted text clipped - 11 lines]> and h must be empty, which means the domain of both is empty, and g> and h are both the empty function from the empty set to itself.

Sure. My main (subconcious) concern was, that the statement is stilltrue, for topological spaces and continuous maps instead of sets and maps,and that with just a little variation, Adams proof would catch thesesituations too.

Marc


Adam - 09 Jul 2004 20:16 GMT> Sure. My main (subconcious) concern was, that the statement is still> true, for topological spaces and continuous maps instead of sets and maps,> and that with just a little variation, Adams proof would catch these> situations too.

Could you elaborate on what you mean by topological spaces andcontinuous maps? That seems very interesting.

Thanks, Adam.


William Elliot - 10 Jul 2004 06:38 GMT> Could you elaborate on what you mean by topological spaces and> continuous maps? That seems very interesting.

http://at.yorku.ca/topology/educ.htm


Arturo Magidin - 11 Jul 2004 19:45 GMT[.snip.]

>> But if we have that the domain of f is empty, and in addition we have>> that f o g = f o h for some functions g and h, then the codomain of g[quoted text clipped - 5 lines]>and that with just a little variation, Adams proof would catch these>situations too.

Ah, but the "real reason" that 'one-to-one/injective' is equivalent to'cancellable on the left' for topological spaces (and 'surjective' isequivalent to 'cancellable on the right') is something else: it isthat given any set there is a topological structure on it that makesany map of sets into it continuous (the indiscrete space), and a

topological structure that makes any map of sets that has it as domaincontinuous (the discrete space); respectively.

So you have a good notion of "free topological space in one element",which gives you that monomorphisms (maps which can be cancelled on theleft) must be injective; and the discrete space gives you thatepimorphisms (maps which can be cancelled on the right) must besurjective.

But restrict the situation even a little, say, to Hausdorff spaces,and the proof no longer works: "cancellable on the right" now becomesequivalent to "image is dense", not to 'surjective'. You still have a'free Hausdorff space on one element' which gives you that cancellableon the left is equivalent to injective, but you lose the other clauseof the theorem. And in other concrete natural categories you lose theinjectivity clause as well (e.g., in the category of divisble groups,the quotient map Q -> Q/Z is a monomorphism, so it can be cancelled onthe left, even though it is not injective).

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Adam - 12 Jul 2004 03:18 GMT> Ah, but the "real reason" that 'one-to-one/injective' is equivalent to> 'cancellable on the left' for topological spaces (and 'surjective' is[quoted text clipped - 3 lines]> topological structure that makes any map of sets that has it as domain> continuous (the discrete space); respectively.

All I can say is "wow!" I hope to be able to understand that in thefuture. It just shows me how interesting mathematics is as one progressesfurther with it. I'm trying to learn what different mathematical mean orrepresent, and then how they are formally defined; that is, I'm trying tolearn the conceptual metaphors that mathematicians think with.Day by day,I'll increase my understanding. It's great motivation to read such posts.

Take care, Adam.


Arturo Magidin - 12 Jul 2004 19:01 GMT>> Ah, but the "real reason" that 'one-to-one/injective' is equivalent to>> 'cancellable on the left' for topological spaces (and 'surjective' is[quoted text clipped - 6 lines]> All I can say is "wow!" I hope to be able to understand that in the>future.

(-:

There are two things at work in that paragraph there: the obvious oneis topology. This is a nice subject (and, in its "point-set" version,something you can try looking at once you finish with set theory; butit grows increasingly complicated as it acquires other 'surnames' like"algebraic topology" and so on...).

Another thing lurking in the background is something called "categorytheory", also known affectionately and derisively (depending on whoyou ask) as "abstract nonsense". Not everyone's cup of tea, though asmall splash of it here and there can be very useful.

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Thomas Heye - 09 Jul 2004 06:06 GMTHi Marc,but what would you gain by allowing empty sets? After all, what is the goodof having functions from an empty set to another? How could one define afunction froma set A to an empty set? Taking the abstract notion of afunction from a set A to a set B, then f is a subset of the ordered pairs(a,b), where a in A and b in B. -- But of course you're right: The factitself is not used in the proof.Regards,Thomas

> > Hi,> >[quoted text clipped - 41 lines]>> Marc


Marc Olschok - 09 Jul 2004 18:05 GMT> Hi Marc,> but what would you gain by allowing empty sets? After all, what is the good[quoted text clipped - 3 lines]> (a,b), where a in A and b in B. -- But of course you're right: The fact> itself is not used in the proof.

Actually, the statement of Part (ii)

>> > Theorem 4.4.4. Let A and B be non-empty sets, and let f: A -> B be a>> > function.>> > (ii) The function f is injective iff f o g = f o h implies g = h for all>> > functions g, h: Y -> A, for all sets Y.

does not depend on A or B being nonempty, the result is true withoutthis restriction. Hence I was just curious if perhaps some other partof the theorem would depend on it.

Adams first part of the proof did use A nonempty in producinga left inverse for f.Of course in the context of the above wording this is o.k.

Marc


Adam - 09 Jul 2004 20:14 GMT> does not depend on A or B being nonempty, the result is true without> this restriction. Hence I was just curious if perhaps some other part[quoted text clipped - 3 lines]> a left inverse for f.> Of course in the context of the above wording this is o.k.

Hi,

Your question in regards to why the theorem concerns non-empty sets isvalid. The author of my book tends to write similar theorems as parts of asingle theorem, for which any needed property is stated for the whole. Forexample, if part (i) of a theorem requires a set to be empty, but part (ii)does not, he states the requirement for both parts in order to simplifythings. However, as he states, his main motivation for requiring some setsto be empty is to not have to deal with the trivial cases, as a number ofpeople have commented on.

This the entire theorem as it is presented in the book; part (i) wassolved as an example by the author.

Theorem 4.4.4. Let A and B be non-empty sets, and let f: A -> B be afunction.(i) The function f is surjective iff g o f = h o f implies g = h for allfunction g, h: B -> X, for all sets X.(ii) The function f is injective iff f o g = f o h implies g = h for allfunctions g, h: Y -> A, for all sets Y.

Thanks, Adam.


Marc Olschok - 10 Jul 2004 18:41 GMT>> does not depend on A or B being nonempty, the result is true without>> this restriction. Hence I was just curious if perhaps some other part>> of the theorem would depend on it.

>[...] > This the entire theorem as it is presented in the book; part (i) was[quoted text clipped - 6 lines]> (ii) The function f is injective iff f o g = f o h implies g = h for all> functions g, h: Y -> A, for all sets Y.

Well, nothing in the theorem needs A or B nonempty. The only relatedsituation, where empty sets may cause trouble is in the statement"every injective map between to sets has a left inverse", becausefor a map emptyset ---> X with X nonempty, this does not hold;but as mentioned before, this stronger property is not needed.

The author of the text might have a valid point in trying to excludetrivial cases, but still it might cause problems later on; suppose,that at some later stage he _needs_ to use the statement above withoutknowing if the sets in question are nonempty? Strictly speaking, he couldnot invoke the theorem at that point without dealing with special casesad hoc.

Marc


Adam - 09 Jul 2004 20:19 GMT> How exactly, did the text define "injective function" ?

Definition. Let f: A -> B be a function.(1) The map f is injective (also known as one-to-one or monic) if x != yimplies f(x) != f(y) for all x, y in A; equivalently, if f(x) = f(y) impliesx = y for all x, y in A.(2) The map f is surjective (also known as onto or epic) if for every b

in B, there exists some a in A such that f(a) = b; equivalently, if f_*(A) =B.(3) The map f is bijective if it is both injective and surjective.

Thanks, Adam.


David C. Ullrich - 08 Jul 2004 16:13 GMT>Hi,>[quoted text clipped - 3 lines]>comments about. It is a simple theorem, but I am just learning set theory,>and so it is a good theorem for me to prove.

Seems exactly right. (Many versions of the proof would be a lot shorter - for that matter it's the sort of thing that a "grownup"mathematician wouldn't bother proving at all. But if we wantto include all the details, which is a good thing right now,then it's just about right.)

>Theorem 4.4.4. Let A and B be non-empty sets, and let f: A -> B be a>function.[quoted text clipped - 14 lines]>>Thank you, Adam.

************************

David C. Ullrich


Adam - 09 Jul 2004 20:47 GMT> Seems exactly right. (Many versions of the proof would be a lot> shorter - for that matter it's the sort of thing that a "grownup"> mathematician wouldn't bother proving at all. But if we want> to include all the details, which is a good thing right now,> then it's just about right.)

Hi,

Thanks for checking my proof. I understand that these are simply proofs,but someone has to begin learning mathematics somewhere. It's not likepeople awake one morning and fully understand abstract algebra or matrixtheory. I'm quite humble at this point in my study of mathematics, as I knowthere is an enormous amouth about which I know nothing. It is difficult attimes to motivate myself to study mathematics, so some days I just dont readanything. However, I enjoy what I am studying so much that I will spend myfree time learning it instead of watching T.V. now. The path that I have setfor myself is to finish Proof and Fundamentals: A first course in abstract

mathematics, and also complete all 400+ exercises. After that I will knowset theory, basics of group theory and lattices, and the Peano Postulates.Then I will proceed to learn linear algebra, but I will need a differenttext than the one I have because I wish for their to be more rigor. Anonline text that I have been reading, and that seems extremely good, ishttp://www.math.miami.edu/~ec/book/book.pdf. When I can understand thatbook, I will be very proud of myself. Then I will learn real analysis,complex analysis, PDE's, combinatorics, probability theory, and some morepure mathematics that I discover along the way. Hopefully then mymathamatics will be suffciently mature to put it to good use.

Thank you for responding, Adam.


William Elliot - 10 Jul 2004 06:44 GMT> http://www.math.miami.edu/~ec/book/book.pdf. When I can understand that> book, I will be very proud of myself. Then I will learn real analysis,> complex analysis, PDE's, combinatorics, probability theory, and some more> pure mathematics that I discover along the way.

Math is infinite, mind finite.

injective and surjective

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