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Homework Help: Linear algebra homework

  1. Nov 13, 2005 #1
    I'm really stuck with my homework - it seems to be easy, but...
    So the first one:
    Find the most natural bijection between these two sets:
    [tex](X \times Y)^Z , X^Z \times Y^Z [/tex]

    The second thing I'm stuck with:
    Proof for arbitrary [tex]f: X \rightarrow Y , g: Y \rightarrow Z[/tex] and sets:
    [tex]A \subseteq X , B \subseteq Z[/tex] :
    [tex](g \circ f)^{-1} (B) = f^{-1}(g^{-1}(B))[/tex]

    And the last one:
    Let [tex]f: X \rightarrow Y [/tex] be an arbitrary function. Proof that for every [tex]A,B \subseteq X ; C,D \subseteq Y[/tex]:
    a) [tex]C \subseteq D \Rightarrow f^{-1}(C) \subseteq f^{-1}(D)[/tex]
    b) [tex]f(f^{-1}(C)) \subseteq C[/tex]
    Last edited: Nov 13, 2005
  2. jcsd
  3. Nov 13, 2005 #2


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    Each of those looks to me to be reasonably straight forward. What have you done so far? What are the relevant definitions?
  4. Nov 13, 2005 #3
    Well in the first one I don't really know what kind of answer they are expecting.

    I found out that the second one is really easy.

    In the third one I don't understand why there is inclusion instead of equality (because an inverse function exist only when the original function is bijective, therefore in b) there should be equality I think)
    Last edited: Nov 13, 2005
  5. Nov 14, 2005 #4
    can anyone help me at least with the first one?
    Please :frown:
  6. Nov 14, 2005 #5


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    For (1):

    If someone said "Please write down a bijection [itex](X \times Y)^Z \rightarrow X^Z \times Y^Z [/itex]", what bijection would you write down? That is almost certainly the one they're asking for.

    For (2):

    For any binary relation [itex]f[/itex] (in particular, a function is a binary relation), the relation [itex]f^{-1}[/itex] is defined. (But is not usually a function)

    Recall that for any binary relation R on (S, T), we can define, for [itex]A \subseteq S[/itex]:

    R(A) := \{ t \in T \, | \, \exists s \in A: s \, R \, t \}

    (The notation for this isn't really standard -- for example, I would really prefer to write it as [itex]A \cdot R[/itex], maybe without the dot)

    For a function [itex]f : S \rightarrow T[/itex], recall that [itex]f[/itex] is merely a binary relation on (S, T). Thus, we obtain the direct image of a subset A of S:

    f(A) = \{ t \in T \, | \, \exists s \in A : f(s) = t \}

    or, more simply,

    f(A) = \{ f(a) \, | \, a \in A \}

    The inverse relation just does things in the opposite order. [itex]t \, R^{-1} \, s[/itex] if and only if [itex]s \, R \, t[/itex]. For [itex]f[/itex], Plugging into the definition, we obtain the inverse image of a subset B of T:

    f^{-1}(B) = \{ s \in S \, | \, \exists t \in B : t \, f^{-1} \, s \}
    = \cdots = \{ s \in S \, | \, f(s) \in B \}
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