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Thread: Composition of Maps

  1. #1
    Apr 2008

    Composition of Maps

    Let A,B be sets and let f:A \rightarrow B and g,h:B \rightarrow A be functions.

    (1) Suppose h o f is an injective map from A to itself. Show that f is injective.

    (2) Suppose now that f o g = 1_{B} and hof = 1_{A}. Show that f is bijective and g=h.

    My attempt:

    (1) h o f is 1-1 \Leftrightarrow \forall a,a' \in A such that a \neq a', h o f(a) \neq h o f(a')

    \Leftrightarrow h(f(a)) \neq h(f(a'))

    \Leftrightarrow f(a) \neq f(a')

    It follows that a \neq a'. Is this proof correct? Do I need to add any more explanation?

    (2) I need some help with question two, so I can get started. Btw, the 1_A and 1_B notation represents the identity.

    Thanks in advance,
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  2. #2
    Junior Member
    Aug 2009
    Proof of (1) is basically correct except for some typos:
    Quote Originally Posted by Roam View Post
    It follows that a \neq a'
    This is where you began, what you proved is that f(a) \neq f(a'). Also,  h(f(a)) \neq h(f(a')) is not equivalent to
     f(a) \neq f(a'), but the implication from left to right holds, and that is what you need.

    (2) f is injective: apply (1) to h \circ f = 1_A is injective.
    f is surjective: if it is not, then f \circ g also isn't surjective, but f \circ g = 1_B is surjective.
    Further, we have g= 1_A \circ g=h \circ f \circ g=h \circ 1_B=h, because composition of functions is associative.
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