Let f: X rightarrow Y be a map and A_1, A_2 Subset X. Prove that f(A_1Intersecti

ID: 3113257 • Letter: L

Question

Let f: X rightarrow Y be a map and A_1, A_2 Subset X. Prove that f(A_1Intersection A_2) Subset f(A_1)Intersection f(A_2). Show by example that the equality might fail. Let f: X rightarrow Y be a map. For a subset B of Y define the preimage of B (which is a subset of X) by f^-1(B) = {x elementof X| f(x) elementof B}. (Do not confuse the notation for the preimage with the inverse mapping: we are not saying that f is invertible, we are simply defining a subset of X which we call the preimage of a subset of Y. The reason this notation looks identical to the inverse mapping notation is that if f happens to be invertible, then the preimage of B is the same as the image of B under the inverse mapping of f: think about it for a minute.) Prove that if B_1, and B_2 are subsets of Y. then f^-1(B_1 Intersection B_2) = f^-1(B_1) Intersection f^-1(B_2). Contrast this result with the result in problem 1. Let f: A rightarrow B and g: B rightarrow C be two maps. Show that if g compositefunction f is injective, then so is f. Determine which of the following relations on R are equivalence relations. Provide complete explanations for every relation, i.e. check all three properties of an equivalence relation for each of the relations below. (a) a - b if a - b lessthanorequalto 0: (b) a - b if |a| = |b|: (c) a - b if ab > 0: (d) a - b if |a - b| lessthanorequalto1: (e) a - b if a - b euro Z. Let f: A rightarrow B be a map. Define a relation - on A by x - y if f(x) = f(y). Prove that - is an equivalence relation on A and describe its equivalence classes.

Explanation / Answer

solution:

I am struggling to prove this map statement on sets.

The statement is:

Let f:XYf:XY be a map.

i) A,BX:f(AB)=f(A)f(B)A,BX:f(AB)=f(A)f(B)
ii) A,BX:f(AB)f(A)f(B)A,BX:f(AB)f(A)f(B)
iii) ff is injective A,BX:f(AB)=f(A)f(B)A,BX:f(AB)=f(A)f(B)

My problem is: I know how to operate on sets, I know how to operate on sets, but I don't know how and where to start the proof, the biggest problem in mathematics, I think.

2.The function g : Y X is said to be a right inverse of the function f : X Y if f(g(y)) = y for every y in Y (g can be undone by f). In other words, g is a right inverse of f if the composition f o g of g and f in that order is the identity function on the domain Y of g. The function g need not be a complete inverse of f because the composition in the other order, g o f, may not be the identity function on the domain X of f. In other words, f can undo or "reverse" g, but cannot necessarily be reversed by it.

Every function with a right inverse is necessarily a surjection. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice.

If f : X Y is surjective and B is a subset of Y, then f(f 1(B)) = B. Thus, B can be recovered from its preimage f 1(B).

For example, in the first illustration, above, there is some function g such that g(C) = 4. There is also some function f such that f(4) = C. It doesn't matter that g(C) can also equal 3; it only matters that f "reverses" g.

3.given that gof is a surjectve that each element in c. pre image in A under the map gof is g E c then three exist on xEA such that (gof)x=y it g(f(x)=y so g(b)=y for always B susetof b for each element subset of c. G is on to.

thats is g(f(x1)=g(f(x2))

(gof)(x1)=(gof)(x2) there fore x1=x2

f is one one g:f is injective

as for chegg rule only answer 3 questions all the best

favorite