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Let $f:V\to W$ be a linear map of finite-dimensional vector spaces. By simply counting dimensions and using rank-nullity, it is clear that $V\cong \mathrm{im}\,f\oplus\mathrm{ker}\,f$. I want to know if this holds on general vector spaces.

In fact, the first isomorphism theorem tells us that $\mathrm{im}\,f\cong V/\mathrm{ker}\,f$. Now consider $V/\mathrm{ker}\,f\oplus \mathrm{ker}\,f$. For every equivalence class in $V/\mathrm{ker}\,f$, fix a representative. Then $([x],k)\mapsto x+k$ is a bijection, so it seems $V\cong V/\mathrm{ker}\,f\oplus \mathrm{ker}\,f\cong \mathrm{im}\,f\oplus\mathrm{ker}\,$.

In order to pick a representative in each class, it seems Choice is required to find a choice function on $V/\mathrm{ker}\,f$. Does this require the full strength of Choice, or is it weaker?

Ryan Unger
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  • what do you mean by $\text{im} \ f \oplus \text{ker}$? – Sidharth Ghoshal Jul 24 '16 at 07:18
  • Do you mean to say $V/\text{ker} \ f \oplus \text{ker} \ f \cong \text{im} \ f $? – Sidharth Ghoshal Jul 24 '16 at 07:20
  • According to Asaf from a post three years ago (link), the axiom of choice is equivalent to subspaces having direct complements. – anon Jul 24 '16 at 07:20
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    @frogeyedpeas OP means img+ker as an external direct sum presumably. – anon Jul 24 '16 at 07:21
  • @arctictern: Well, not according to me. More in accordance with general literature about the axiom of choice and vector spaces. – Asaf Karagila Jul 24 '16 at 07:21
  • Speak of the devil. Well, I am not familiar with the general literature, and you were the first thing I found from google. As the comments in the link suggested, maybe you really are Freddy Kruger. :-) – anon Jul 24 '16 at 07:21
  • @arctictern: At present time I cannot confirm nor deny any knowledge of involvement in such activities. – Asaf Karagila Jul 24 '16 at 07:34
  • Note that $([x], k) \mapsto x + k$ is a bijection, but not necessarily a linear transformation! However, the title of this post is true. See https://math.stackexchange.com/a/908233/711748. – jskattt797 Nov 07 '20 at 06:16

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Yes, this requires the full strength of the axiom of choice. The proof goes through the following theorem.

The following are equivalent of $\sf ZF$.

  1. The axiom of choice.
  2. If $V$ is a vector space, and $W\subseteq V$, then $W$ has a direct complement in $V$.

Now it's easy, because given any $W$, consider the obvious map from $V$ to $V/W$, $V\simeq W\oplus V/W$, this means that $W$ has a direct complement. Therefore the axiom of choice must hold.

In particular, if the axiom of choice fails, then $V\simeq\ker f\oplus\operatorname{im} f$ must fail for some $V$ and $f$.

Community
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Asaf Karagila
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  • Would countable or dependable choice be enough if we restrict the dimensionality of $V$? If yes, how much do we have to restrict it (e.g. countable choice for countable dimension)? – yuggib Jul 24 '16 at 09:38
  • Yeah, that sounds about right. You want to choose fibers for the quotient map. So you probably need something like countable choice if the image has a countable dimension. – Asaf Karagila Jul 24 '16 at 09:56