Question

Let A be a finite set and let f be a surjection from A to itself. Show that f is an injection.

Use Theorem 1, 2 and corollary 1.

Theorem 1 : Let B be a finite set and let f be a function on B. Then f has a right inverse. In other words, there is a function g: A->B, where A=f[B], such that for each x in A, we have f(g(x)) = x.

Theorem 2: A right inverse for a function is an injection

Corollary 1: An injection from a finite set to itself is a surjection

Answer #1

Let f be a function from the set of students in a discrete
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(a) Under what conditions is f an injection?
(b) Under what conditions is f a surjection?
(please show all work)

Let A be a finite set and f a function from A to A.
Prove That f is one-to-one if and only if f is onto.

Let f be a function which maps from the quaternion group, Q, to
itself by f (x) = i ∙x, for i∈ Q and each element x in Q. Show all
work and explain! (i) Is ? a homomorphism? (ii) Is ? a 1-1
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Let (G,·) be a finite group, and let S be a set with the same
cardinality as G. Then there is a bijection μ:S→G . We can give a
group structure to S by defining a binary operation *on S, as
follows. For x,y∈ S, define x*y=z where z∈S such that μ(z) =
g_{1}·g_{2}, where μ(x)=g_{1} and μ(y)=g_{2}. First prove that
(S,*) is a group. Then, what can you say about the bijection μ?

Let (G,·) be a finite group, and let S be a set with the same
cardinality as G. Then there is a bijection μ:S→G . We can give a
group structure to S by defining a binary operation *on S, as
follows. For x,y∈ S, define x*y=z where z∈S such that μ(z) =
g_{1}·g_{2}, where μ(x)=g_{1} and μ(y)=g_{2}.
First prove that (S,*) is a group.
Then, what can you say about the bijection μ?

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: B → N|B| , you may consider for instance the function h : A ∪ B →
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Let F be the set of all finite languages over alphabet {0, 1}.
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Proposition 16.4 Let S be a non–empty finite set.
(a) There is a unique n 2 N1 such that there is a 1–1
correspondence from {1, 2,...,n} to S.
We write |S| = n. Also, we write |;| = 0.
(b) If B is a set and f : B ! S is a 1–1 correspondence, then B is
finite and |B| = |S|.
(c) If T is a proper subset of S, then T is finite and |T| <...

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set B to set C. Assume that f °g is one-to-one and function f is
one-to-one. Using proof by contradiction, prove that function g
must also be one-to-one (in all cases).

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