Talk:Surjective function

Origin of terminology?

Anyone know where the terminology "onto" comes from? Listing the etymology of the term may help explain the concept.

Page rename

For consistency with injective function perhaps this page should be renamed surjective function and a redirect sent there from here.

Sounds good -- who wants to do it?

I did it. --Soapergem 09:30, 4 May 2006 (UTC)

You forgot to move the talk page (I did it). -lethe ^{talk +} 13:51, 4 May 2006 (UTC)

Picture

Why is the picture a diagram of a non-surjective function? This is a bit confusing. —Preceding unsigned comment added by 129.31.68.80 (talk) 12:27, 3 February 2011 (UTC)

Diagram of non surjective function has two points in Y labeled y₁ and is missing y₃. The upper point Y₁ appears to actually be the missing Y₃. 173.184.65.10 (talk) 19:27, 25 September 2012 (UTC)

Reversible and Inverse

if a surjective function is revrsible, how does the first picture there represent a reversible function. The C appears to be mapped to by two elements of the domain; wouldn't that make it no surjective? anton

It's not very clear in the article. "Reversible" simply means that if g(C) = 4, f(4) = C. It doesn't matter that g(C) can also equal 3. I'm going to try to put that into the article. --KSnortum (talk) 20:52, 25 November 2007 (UTC)

There seems to be a contradiction in the part "surjection and epimorphism". First it says that there is equivalence surjection and epimorphism , then it says that it holds only for the category of sets!! — Preceding unsigned comment added by 192.76.172.10 (talk) 13:13, 26 August 2013 (UTC)

Rendering

This page isn't rendering correctly for me in Firefox 2.0.0.9 and IE 7.0.5730ish. Unfortunately, the preview while editing looks fine and I don't know what's causing the problem, so I may have to post a number of edits to get it right, or better. --KSnortum (talk) 18:32, 25 November 2007 (UTC)

Got a brilliant idea and decided to do this work in a subpage instead of the live article. I'll copy it over if and when I get it working. I'll try to keep track of changes here but if I make a regression error please forgive me (and correct it). --KSnortum (talk) 18:53, 25 November 2007 (UTC)

I put the images in a table; that helps render them in Firefox and IE. I think it looks okay but feel free to improve it. --KSnortum (talk) 19:59, 25 November 2007 (UTC)

Universal property of quotients

I have never commented on a talk-page before so I am sorry if I am breaking some rules (please tell me if this is the case so I can improve for next time :) !). In the "Other properties" section you talk about the universal property of quotient spaces - in particular you say that for any h:X->Z there is an surjective function f:X->Y and an injective function g:Y->Z such that g composed with f is equal to h. I don't think this is quite correct. I believe an extra condition is required that h(x)=h(x') implies that f(x)=f(x'). This puts a condition on the kind of quotient spaces we are allowed (nothing is free in life! :) ). Here is my reasoning:

We construct g(y) by picking a point x in f^{-1}(y) and letting g(y)=h(x). There is no guarantee then that when we decide what g(y') is equal to i.e. pick x' in f^{-1}(y') such that g(y')=h(x') that h(x') does not equal h(x). In particular it could be that g(y)=g(y') without y=y'.

Here is a specific example: - let f:{1,2} -> {[1],[2]}, in other words f splits {1,2} into odds and evens
- let h:{1,2} -> {1}
- Now construct g[1]. f^{-1}[1]=1. Hence g[1]=h(1)=1.
- Now construct g[2]. f^{-1}[2]=2. Hence g[2]=h(2)=1.
- Hence even though [1] is not equal to [2], g[1]=g[2]. So g is not injective.

However if we include the condition I stated we have the following proof that g is injective:
- We already know that f is surjective as it is the canonical map between the space X and its quotient space Y
- If g(y)=g(y') then, because f is surjective there exists x and x' in X such that f(x)=y and f(x')=y'.
- g has the special property that g \circ f = h. Therefore h(x) = g \circ f(x) = g \circ f(x') = h(x').
- The additional condition now demands that f(x)=f(x'). Hence y=y'.
So g is injective.

Hope this is correct. Db1685 (talk) 09:16, 2 September 2008 (UTC)

I don't understand your objection. The article says that given h we can choose f and g with the given properties. You seem to have read it as saying that given f and h we can choose g with the given properties. In that case, yes, you need more conditions on f than that it be surjective, but that case is not what is in the article. Since you asked, you did break one of the rules of Wikipedia talk pages, which is that new discussions should go at the bottom of the page in a new section. You can easily do this by clicking the 'new section' tab at the top of the page. (of course, replies to existing discussions should go in the same section, preferably indented with colons, like this one.) Algebraist 09:35, 2 September 2008 (UTC)

I think there has been a misunderstanding here. Given h:X->Y if we want f surjective and g injective we still need f to have the property I mentioned. I can see now how it could be read the way you stated. However, with all due respect, I feel the way it is written is a little misleading. I agree with the first sentence. But then you say:

"To see this, define Y to be the sets h^{−1}(z) where z is in Z. These sets are disjoint and partition X. Then f carries each x to the element of Y which contains it"

So now you have defined a surjective f using only h. I agree with this. f is exactly the canonical map between the X and the quotient space Y. You then proceed to define g:

"and g carries each element of Y to the point in Z to which h sends its points."

I understand by no means is this a proof but, to me, it suggests that you can just construct g as I discussed above without imposing the condition I stated upon it. In particular the statement:

"g is injective by definition"

is the part I found troubling. It is not injective unless the extra condition is imposed and a definition without that condition is possible as I illustrated. At least - this is how I read it.

My apologies for being a pedant. Db1685 (talk) 10:02, 2 September 2008 (UTC)

P.S. Thanks for the tip about posting! I will remember next time :).

g is defined to be the function from the set $\{h^{-1}(z)|z\in Z\}$ (the empty set excluded) to Z that maps $h^{-1}(z)$ to z. This function is indeed injective by definition. I really don't see what the problem is here. Algebraist 10:12, 2 September 2008 (UTC)

Again I see what you meant. However consider the fact that h^{-1}(z) need not be a single point. Hence g would map many points to z. So it could not be injective. This is exactly what happens in the example I gave. —Preceding unsigned comment added by Db1685 (talk • contribs) 10:23, 2 September 2008 (UTC)

No, g does not map many points to z. g maps the single point $h^{-1}(z)$ of the set $\{h^{-1}(z)|z\in Z\}$ to z. In the example you gave, with h mapping {1,2} to {1}, the construction gives Y={{1,2}}, a one-point set, and g is the unique function from {{1,2}} to {1}. The problem with your example is that you have chosen f and Y arbitrarily, rather than constructing them as given in the article. Algebraist 10:28, 2 September 2008 (UTC)

Yes, I see your point. You were right and I got my f,g,h muddled up. I was using f(X) as the quotient space of X. I was proving something slightly different. Thanks for clearing that up. Db1685 (talk) 10:47, 2 September 2008 (UTC)

Range equal to co-domain

The initial statement in the article that the range of a surjective function must be equal to its co-domain seems simply to be false: clearly, the co-domain can (a) be smaller than the domain and (b) contain elements which are not in the domain. It's therefore not clear what sense of equality is meant here. But I'm not a mathematician so I just wanted to check if I'm confused about this. Cadr (talk) 14:24, 20 November 2009 (UTC)

The article is correct, so yes you are confused. Paul August ☎ 15:04, 20 November 2009 (UTC)

Oh, I read "domain" for "range" for some reason, my mistake. No need for you to be an asshole about it though :) Cadr (talk) 16:58, 24 November 2009 (UTC)

Please accept my apology. I asssure you I didn't mean my comment to be insulting or flip. Although I now see how they might have sounded that way. Sorry for that :( Paul August ☎ 15:26, 25 November 2009 (UTC)

Thanks. I also apologize for my rather flippant reply. Cadr (talk) 15:33, 25 November 2009 (UTC)

No problem. Paul August ☎ 15:37, 25 November 2009 (UTC)

Decomposition

Conversely, if f o g is surjective, then f is surjective (but g need not be).

Shouldn't it be

"g" is surjective (but "f" need not be)?

At least, that's what one of the diagrams on the page illustrates. —Preceding unsigned comment added by 65.110.237.146 (talk) 21:01, 22 September 2010 (UTC)

No, the article is correct. Remember that "f o g" means that g is applied first, then f. --Aleph4 (talk) 19:09, 10 September 2011 (UTC)

Domain

The link to domain go to a disambiguation page. It should be more precise. 216.19.183.88 (talk) 22:25, 10 January 2012 (UTC)

Fixed. Thanks for pointing this out. Jowa fan (talk) 23:25, 10 January 2012 (UTC)

Example of a actual function that is surjective but not bijective with image

I think the images here are exceptionally good. I might add an example of an actual function that is surjective, but not bijective such as  $f(x)=x\cdot sin(x)$

Just noticed that this is a cool example of the product of two odd functions being an even function :) Lfahlberg (talk) 06:27, 29 June 2013 (UTC)

Since nobody objected I added an example (without an image) of a surjective, but not bijective function.Lfahlberg (talk) 15:36, 6 August 2013 (UTC)

Arrow Symbols

I like this page, but would someone please add a table of the different arrow symbols that are commonly used for onto, one-one, etc? It would also be nice to note the LaTeX commands for the symbols. Jfgrcar (talk) 20:02, 2 June 2014 (UTC)