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Math Help - f and dual f

  1. #1
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    f and dual f

    I am preparing for my exam right now and I am solving questions. I proved something but I am not sure. I have uploaded the question with the definitions just in case we have different notations.

    So the questions is:

    V and W are vector spaces. f is a mapping from V to W. f is injective if and only if f dual is surjective.

    (I have solved something really similar to this so this is why I came up with the solution but still I am not sure.)

    The file is more detailed. Appreciate the feedbacks. Thank you.
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  2. #2
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    Re: f and dual f

    I think what you have is correct, but it's a bit hard to follow.

    First, you don't really define $f^{\ast}$. I think what you probably intend is:

    $f^{\ast}:W^{\ast} \to V^{\ast}$ given by, for $\varphi \in W^{\ast}$: $f^{\ast}(\varphi) = \varphi \circ f$.

    Let's see if we can provide a more direct proof:

    Suppose $f$ is injective, and let $\psi$ be any element of $V^{\ast}$.

    Since $f$ is injective, there is $g \in \text{Hom}_{\ F}(\text{im }f,V)$ with $g \circ f = 1_V$.

    Now consider $g^{\ast}:V^{\ast} \to (\text{im }f)^{\ast}$.

    By definition: $(f^{\ast}\circ g^{\ast})(\psi) = f^{\ast}(g^{\ast}(\psi)) = (g^{\ast}(\psi)) \circ f = (\psi \circ g) \circ f = \psi \circ (g \circ f) = \psi \circ 1_V = \psi$.

    Since $g^{\ast}(\psi) \in (\text{im }f)^{\ast} \subseteq W^{\ast}$, it follows that $g^{\ast}(\psi)$ is an element $\varphi \in W^{\ast}$ with $f^{\ast}(\varphi) = \psi$, that is, $f^{\ast}$ is surjective.

    On the other hand, suppose $f^{\ast}$ is surjective.

    This, in turn, means that there is $g^{\ast}: V^{\ast} \to W^{\ast}$ such that $f^{\ast} \circ g^{\ast} = 1_{V^{\ast}}$.

    So for any $\psi \in V^{\ast}$, we have:

    $\psi = 1_{V^{\ast}}(\psi) = (f^{\ast} \circ g^{\ast})(\psi) = f^{\ast}(g^{\ast}(\psi)) = (g^{\ast}(\psi)) \circ f = (\psi \circ g) \circ f = \psi \circ (g \circ f)$.

    In particular, for any $v \in V$, with $\psi = \langle v,-\rangle$, we have $\langle v,-\rangle = \langle (g \circ f)(v),-\rangle$ so that:

    $\langle v - (g \circ f)(v),x\rangle = 0$ for all $x \in V$, including $x = v - (g \circ f)(v)$, so that $v = (g \circ f)(v)$ for all $v \in V$, that is $g \circ f = 1_V$, and $f$ is injective.
    Thanks from davidciprut
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  3. #3
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    Re: f and dual f

    Thank you, I really appreciate the feedback.
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