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So to show exactness at Hom$(V'',W)$, I need to show $g$# is injective.
To show exactness at Hom$(V,W)$, I need to show Im(g#) = Ker(f#), which means f# sends all of Hom(V,W) to 0?

I'm unable to see the big picture though the symmetry clearly should mean something. Please help me. Thank you.

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It is not clear to me what you are asking for. Do you want concrete help with this exercise, that is how to prove the exactness of the hom-sequence, or do you want to get a better understanding of exactness and how to interpret it? Also, exactness at $\operatorname{Hom} (V,W)$ does mean that $\operatorname{img} (g\#) = \ker (f\#)$ and this does not mean that $f\#$ sends all of $\operatorname{Hom}(V,W)$ to $0$, but only the image of $g$. –  k.stm Feb 17 '13 at 15:18
    
I would like to know both. –  mez Feb 17 '13 at 15:21
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How to prove exactness of the hom-sequence:

We have to do two things:

  1. Show $\operatorname{img} (0) = \ker (g\#)$, i.e. $g\#$ is injective, as you said.
  2. Show $\operatorname{img} (g\#) = \ker (f\#)$, which can be split up to showing:

    (a) $\operatorname{img} (g\#) \subseteq \ker (f\#)$, and

    (b) $\operatorname{img} (g\#) \supseteq \ker (f\#)$.

Proving $g\#$ has trivial kernel, take an $γ ∈ \ker (g\#)$, wich is a linear map $V'' → W$. Now, $g\#(γ) = 0$, this means $γ∘g = 0$. Use the surjectivity of $g$.

Proving $\operatorname{img} (g\#) \subseteq \ker (f\#)$, take $β ∈ \operatorname{img} (g\#)$. It can be written as $β = g\# (γ) = γ∘g$ with $γ ∈ \operatorname{Hom} (V',W)$. What happens if you $f\#$ this $β$ now?

Proving $\ker (f\#) \subseteq \operatorname{img} (g\#)$ is the trickiest bit. Take $β ∈ \ker (f\#)$, meaning $β∘f = 0$. So you know $\ker (g) = \operatorname{img} (f) \subseteq \ker (β)$. This should give you a hint that it’s possible to find a $γ \colon V'' → W$ such that $β = γ∘g$. Try to construct such, using the surjectivity of $g$.

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