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Suppose that V and W are vector spaces (say over the reals) and that T is a linear surjection from V onto W with kernel K. By picking a basis for W (using choice) and pulling it back through T to a linearly independent set in V (again using choice) it is not difficult to get a linear embedding i of W into V satisfying T(i(w)) = w. It is then easily checked that the map $(k,w) \mapsto k + i(w)$ gives an isomorphism from $K \oplus W$ onto V. This all seems a little hamfisted though for such an algebraic result. I thought I would check whether there is some way to get such an isomorphism which avoids choice. I would guess there is no chance since there this isomorphism is so devastatingly nonunique... In fact I'm not sure I even want to ask this question anymore now that I think about it...

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The choice-free result is that $T$ induces an isomorphism from the quotient space $V/K$ to $W$. –  Chris Eagle Mar 17 '11 at 22:49

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up vote 4 down vote accepted

You do need choice in general:

without it, there may exist vector spaces none of whose nontrivial subspaces has a complement (Herrlich, Axiom of Choice, LNM 1876, Disaster 4.43).

-JS Milne, MathOverflow, Dec 1, 2009

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Saints preserve us! Thanks for the reply –  Mike F Mar 17 '11 at 22:55

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