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Problem 3 We let f:X _ Y be a function, prove that f is injective if and only if f(An B) = f(A) n f(B) for all A,B = X. (@Hint: To prove = (i.e: f(An B) = f(A) n f(...

Question

Problem 3 We let f:X _ Y be a function, prove that f is injective if and only if f(An B) = f(A) n f(B) for all A,B = X. (@Hint: To prove = (i.e: f(An B) = f(A) n f(B) implies f is injective) part, you can consider "proof by contradiction" and derive a contradiction by considering suitable choices of Aand B):

Problem 3 We let f:X _ Y be a function, prove that f is injective if and only if f(An B) = f(A) n f(B) for all A,B = X. (@Hint: To prove = (i.e: f(An B) = f(A) n f(B) implies f is injective) part, you can consider "proof by contradiction" and derive a contradiction by considering suitable choices of Aand B):



Answers

Suppose that $g$ is a function from $A$ to $B$ and $f$ is a function from $B$ to $C .$ Prove each of these statements.
a) If $f \circ g$ is onto, then $f$ must also be onto.
b) If $f \circ g$ is one-to-one, then $g$ must also be one-to-one.
c) If $f \circ g$ is a bijection, then $g$ is onto if and only if $f$ is one-to-one.

In this problem of relations and functions we have given that A and B are two sets. Yeah, we have to show their the council which is from a cross B two, be across a such a day function, F A B is equal to be, It is by objective, so a function is said to be objective and then if it is 11 and onto concerned. So first we have to check for on two fronts 11 function or we can say interactivity. So first, the step of the way, check for 11 function 11 quantum. So the answer is said to be 11 if head even and say B one which belongs to here A Crosby and say 82 and B two, which also belongs from a cross big such that the function here here F off even and we won is equals two F of a two and B two. And this implies that everyone is equal to 82 beaver is equal to be too. If this condition is satisfied, we say that the sponsor is 11 So from here, F F A N B one is equal to say be even and even and F of it'll be too is B two and a two. So any order pair is said to be equal if B one is equal to be two and a one is equal to a two since this condition is satisfied. So we can say that the function is 11 and now we have to check for onto johnson. So here we are taking for onto function. So step two. Mhm. Check for check four subjectivity or onto once. Um Mhm. So let's be and with the element of be close. So this is here B and which belongs to be across A then head, element of which belongs to the given set A an element or B. Which belongs to the given said B. And this implies that here baby A. And we would be belongings from across B. So this will be from Crosby for all values of B. And A. So we say that for all values of the order will be and A. Which belongs to the big cross trade did exist. So there exist ordered pair. The ordered pair would be A. And B. A. And B. Which belongs to air Crosby. So that means every pair would have its image. Or we can say extremity in could've been. So that's why we say that the function F. Is here on two since the fence anyone went on to. So we can say that hence funds managed by objective. Hence S. Is by objective. So this is the answer.

In this question. The given function is hey folks access, it was to access word minus one into Access were -2 here. Okay. First of all we can see that in this case can be written as x plus one and two X -1 into access with -2. Okay. And their domain is are here. You can see that means the function is defined for all real values. We stay here. If we put here minus one then this provides at minus one plus 10 and if we put the other one then the second fun factor converts into zero. So you can see the minus one and one in order equal here. But you can see value it. Both of them is equal due to this reason. We can say that what is happening here is that Excellent and X two belongs to the domain and are such that they are not equal. But you found the values equal for both of them means we can say here that the function is not 11 or I can also say that it is not inject you here. Okay, sort of the given options. The B option is corrected because it is saying that the function is not injected here. Okay, this is the answer here. Thank you.


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