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Thread: Composition of Maps

  1. #1
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    Composition of Maps

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

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

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


    My attempt:

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

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

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

    It follows that $\displaystyle a \neq a'$. 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 $\displaystyle 1_A$ and $\displaystyle 1_B$ notation represents the identity.

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

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