Sooo I tried doing this the first time and I failed, then I looked online for solutions and I found one that seemed understandable but still I would've done it in a different way, upon trying to do it with this different way, i got the wrong answer. Can you tell me where did I go wrong?
So I get that we have 2 options here, either we reflect 3 elements from A onto 1 element in B or we reflect 2 elements from A to 1 element in B and do that again with another 2 elements of A. So I would write that like number of combinations 3 to 1 + number of combinations 2 to 1 x2
So lets start with the first one. We have like a set of these elements (||| , | , |) and we have to rearrange them, and there are 3 combinations (||| , | , |) or (| , ||| , |) or (| , | , |||) . Now me thinking on like writing it graphically (doodling) i would write _ _ _ and place 3 on the first _ then 2 and then 1, bcs like, there are 3 elements to choose from to put on the first dash then we are left with 2 and then we are left with 1, now that is equal to 6 and there are duplicates like baa and baa bcs a is the same element we don't want that. So i would divide by 3! bcs there are 3 dashes. But apparently you should divide by 2! because we have a set of {||| , |} but we can place 2 of | IDK how you would write that mathematically, maybe calling (||| , | , |) a set at first was wrong but still a set can have duplicates right? So in conclusion we have 3 combinations.
Now we must choose 3 elements from A, 3 dashes again 5*4*3/3! because this time they are different. And we get 10 combinations of choosing 3 elements from A. That would make 3 ways of placing them *10 ways of choosing them, 30 ways but we are not done. We have 2 elements left and there are 2 dashes 2*1=2 then there are 60 combinations in total. But I get confused, why didn't we divide this time, for the first choice we have 2 choices, and for the second we have 1. I mean it's either ab or ba but isnt that kind of the same? I guess there is only 1 choice of choosing them but 2 places where they can go, so that makes two choices.
Now thats 60 for the first part, now we go onto the next one, we have a set of {||, |} and 2 duplicates of ||. There are 3 dashes, 3 options for placing 1 one the first dash, then 2 options and 1 option, but we divide by 2! so we get 3, 3 possible rearrangements. then we have to choose 2 elements from A and we have 2 dashes 5*4/2! is 10 then for the ne
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https://preview.redd.it/77ng53tywuy71.png?width=403&format=png&auto=webp&s=6317107211adfe7a493943d7941278627fc286a1
I am given this function to decide if it surjective or not.
https://preview.redd.it/5xxjul0jb1t71.png?width=278&format=png&auto=webp&s=c8e55bc9e93e65d365678f18c9be889062388583
First I used the squaring method to simplify the function into:
https://preview.redd.it/ck0p5l9tb1t71.png?width=216&format=png&auto=webp&s=fbdbe715307be651db09bfaca30a8a16e10a9841
but from here I don't how to prove surjectivty, it is a nightmare to understand. Do I have to just prove Left side = Right side...? What is the plan?
Hey, a quick question. Why is e^x not surjective? (R->R) I’ve written down that f(ln(y))=y and thought it is, but the answer says it is not surjective.
What is an injective function and what is a surjective function?
could you use analogies?
Could you explain it in a simple way?
what do you mean by "each element" ...?
For natural numbers k and n, let
- Inj(k, n) be the number of injective functions from a set of size k to a set of size n. Explicitly, I(k,n) = n!/(n-k)!.
- Sur(n, k) be the number of surjective functions from a set of size n to a set of size k. That is, Sur(n,k) = k! S(n,k), where S(n,k) is a Stirling number of the second kind.
Prove that whenever k < n, Inj(k, n) ≤ Sur(n, k). For example, when k = 2 and n = 4, we have Inj(2, 4) = 12 and Sur(4,2) = 14.
https://imgur.com/a/CAiN8Me
I'm really confused here. On one hand, I'm being told that functions must pass the vertical line test. On the other, I'm learning about surjective functions. How can a function be surjective, when the very meaning of that is a contradiction to the definition of a function?
Is the function f(x)=2x+3 surjective or injective and how do i do it?
Can anyone tell me how they would approach this problem? I know that for cyclic group of order 2, one homomorphism would be to map numbers to its sign (-1 or 1), but I don’t understand how people came up with this. Is there a logical way to construct this homomorphism?
I do not understand the definitions of injective and surjective functions, could you tell me the explanations that helped you understand these concepts? Thank you!
Could someone please help me with understanding / coming up with a function with a surjective graph in R^3 that I can plot?
I’ve been struggling with the definition of surjectivity in R^3 and want to improve my understanding.
I thought about f(x, y, z)=(x, y^2, z^3 ) but really not sure if this works or is even written correctly...
Any help is appreciated, thank you!!
We may presuppose that exp(x) is strictly increasing and continuous. Is my proof correct?
Proof. Let y be a positive real number and let x=0. If y=1, then exp(0)=1 proves this case. Now let's observe the case that y>1. Because exp(x) is strictly increasing and continuous, we may increase x until exp(x)=y. The same reasoning can be applied for y<1. This proves that for all y there exists some real number x such that y=exp(x).
If the proof is correct, can it be generalized by replacing "exp(x)" for any function? This would then be equivalent to the theorem that every strictly increasing and continous function is surjective.
I'm new to Category Theory and have just been introduced to the definition of Epimorphism: f:A->B is epimorphic iff for all C, for all g_1, g_2 (morphisms from B to C), g_1∘f = g_2∘f implies g_1 = g_2.
It is said that this corresponds to surjective functions in the category Set. ("Set" to my current understanding means when the objects are sets and the morphisms are functions i.e. "Set" does not refer to one category of ALL sets and ALL functions, but just any category of this type.)
I'm skeptical of the claim that all epimorphisms indicate surjective functions in Set. I can see the intuition behind the definition, and that it works in certain instances. I've even looked at a proof on stackexchange, but it defined functions which aren't guaranteed to exist by the axioms of category theory, so I'm not convinced this 'proof' works for all examples.
Consider the following Category:
Objects: A, B
Morphisms: identities, along with f:A->B and the required compositions
It follows that f is an epimorphism, for the only morphism with domain B is id_B, and id_B∘f = id_B∘f and id_B = id_B.
Now we may instantiate this category with sets and functions. Let:
A = 2Z (all even integers)
B = {0, 1}
f:A->B is such that f(x) = 1 if x is even, f(x) = 0 if x is odd.
f is clearly not surjective, but in the category defined above it is classified as epimorphic. The problem is that the category lacks the resources (e.g. other morphisms) with which to sufficiently probe f.
Where does my understanding diverge from the consensus?
A function g : Z → Z is defined as g(n) = |n| + 1. Show that this function is not injective and not surjective.
I have been struggling with this one. Thanks in advance
I think I'm onto something.
Hello!
My question is in regards to Linear Algebra Done Right pg 87 Section 3.D. After establishing that invertibility, infectivity, and subjectivity are equivalent for linear operators on finite dimensional vector spaces (3.69), Axler gives an example (3.70) using the vector space of polynomials with real coefficients.
He says, "Example 3.68 shows that the magic of 3.69 does not apply to the infinite-dimensional vector space P(R). However, each nonzero polynomial q has some degree m. By restricting attention to P_m(R), we can work with a finite-dimensional vector space." For reference, the example he uses this on is the linear operator — Tp = ((x^2+5x+7)p)'' . Here is a picture of the excerpt, https://imgur.com/a/qk2iW45
My question is more a question on technique. By looking at the restrictions on an infinite-dimensional vector space, can we always proceed like this? Or, is this a special case since it is a vector space of polynomials? The latter wouldn't make sense to me, since R^/inf seems isomorphic to P(/R) given you have the map e_j = (0,..,0,1,0,..0) --> x^{j-1}. Overall, I just would like some clarification on this. Thank you!
Hello,
I am attempting to create a python code that will determine if a pair of sets is surjective, but I am having a bit of trouble getting the desired output.
So far this is what I have come up with:
def ontoFuncs(x, y):
for i in x:
for j in y:
print((i,j), end=',')
ontoFuncs({'a', 'b', 'c'}, {1, 2})
My output:
('c', 1),('c', 2),('b', 1),('b', 2),('a', 1),('a', 2),
The example output we were given:
>> fs = all_onto_funcs({'a', 'b', 'c'}, {1, 2})
>> for i in range(len(fs))
...print(fs[i])
...
>>
{(a, 1), (b, 1), (c, 2)}
{(a, 1), (b, 2), (c, 1)}
{(a, 1), (b, 2), (c, 2)}
{(a, 2), (b, 1), (c, 2)}
{(a, 2), (b, 2), (c, 1)}
{(a, 2), (b, 1), (c, 1)}
>> fs_2 = all_onto_funcs({1, 2}, {'a', 'b', 'c'})
>> print(fs)
[]
I'm a bit confused on the sample output.... But I'm wondering if anyone has any hints on how to print the surjective functions? I'm not looking for answers, just some hints that could help me move further :)
Update: I solved this :)
my understanding:
- injective f(a1)=f(a2)
- surjective: all a belongs to A, there exists b belongs to B, f(a)=b
- bijective : injective + surjective
I only know this, but I dont know the meaning/defintion of these 3 things!
https://preview.redd.it/przs1367xuy71.png?width=403&format=png&auto=webp&s=e6eace529771b5fce3ca7a78e524ecae12f01292
What is an injective and a surjective function?
could you use analogies?
could you explain them in a simple way?
What do you mean by "each element ..."?
