Return to Answer

(I completely revamped my answer, the previous version (link) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.)

The answer is no. For projectivity and injectivity. The reason is that the axiom of choice is equivalent to the following statement:

If $V$ is a vector space, and $K$ is a subspace of $V$, then there exists a a direct complement for $K$, that is a subspace $S$ such that $S\cap K=\{0\}$ and $S\cup K$ is a generating set for $V$.

If every vector space is projective, consider the case that $V$ is a vector space, $K$ is a subspace and $T\colon V\to V/K$ is the quotient map. Then there is some $h\colon V/K\to V$ such that $h\circ T$ is the identity function. In particular, this means that $K+\operatorname{im}(h)=V$, but $h$ cannot map any nonzero vector into $K$ so this is a direct sum indeed.

Therefore the statement "Every vector space is projective" implies the axiom of choice.

Similarly for injectivity, take the identity function $K\to V$, then there is a projection map whose kernel is a direct complement for $K$.

I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.

(I completely revamped my answer, the previous version (link) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.)

The answer is no. For projectivity and injectivity. The reason is that the axiom of choice is equivalent to the following statement:

If $V$ is a vector space, and $K$ is a subspace of $V$, then there exists a a direct complement for $K$, that is a subspace $S$ such that $S\cap K=\{0\}$ and $S\cup K$ is a generating set for $V$.

If every vector space is projective, consider the case that $V$ is a vector space, $K$ is a subspace and $T\colon V\to V/K$ is the quotient map. Then there is some $h\colon V/K\to V$ such that $h\circ T$ is the identity function. In particular, this means that $K+\operatorname{im}(h)=V$, but $h$ cannot map any nonzero vector into $K$ so this is a direct sum indeed.

Therefore the statement "Every vector space is projective" implies the axiom of choice.

Similarly for injectivity, take the identity function $K\to V$, then there is a projection map whose kernel is a direct complement for $K$.

I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.

The answer is no. Or at least not necessarily. At least for projectivity.

It is consistent that there is a vector space $V$ over $\Bbb F_2$ (or really any finite field) whose underlying set is amorphous, namely it cannot be partitioned into two infinite subsets.

This vector space is not finitely generated, but every proper subspace has a finite dimension.

Pick any such subspace $V'$ and consider $W=V/V'$. Then it is not very difficult to show that the following holds:

1. $W$ is a vector space whose underlying set is amorphous, and every proper subspace is finitely generated. But $|W|\neq|V|$.
2. There is no non-zero homomorphism from $W$ to $V$.
3. The natural quotient map is surjective from $V$ to $W$.

So $W$ is not projective. This also gives an example of an exact sequence which doesn't split. As Jeremy Rickard observes in the comment, this is also a counterexample to injectivity.

(The above can be done easily over infinite fields too, but then $V$ is not amorphous, just weird, and we might not have $|W|\neq|V|$ in general.

Another generalization is that we might allow subspaces to be not just finitely generated, but generated by a well-ordered set up to some cardinal $\lambda$ (which would mean the subspace is free, of course), but the space itself is not generated by any such set.)

I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.

(I completely revamped my answer, the previous version (link) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.)

The answer is no. For projectivity and injectivity. The reason is that the axiom of choice is equivalent to the following statement:

If $V$ is a vector space, and $K$ is a subspace of $V$, then there exists a a direct complement for $K$, that is a subspace $S$ such that $S\cap K=\{0\}$ and $S\cup K$ is a generating set for $V$.

If every vector space is projective, consider the case that $V$ is a vector space, $K$ is a subspace and $T\colon V\to V/K$ is the quotient map. Then there is some $h\colon V/K\to V$ such that $h\circ T$ is the identity function. In particular, this means that $K+\operatorname{im}(h)=V$, but $h$ cannot map any nonzero vector into $K$ so this is a direct sum indeed.

Therefore the statement "Every vector space is projective" implies the axiom of choice.

Similarly for injectivity, take the identity function $K\to V$, then there is a projection map whose kernel is a direct complement for $K$.

I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.

The answer is no. Or at least not necessarily. At least for projectivity.

It is consistent that there is a vector space $V$ over $\Bbb F_2$ (or really any finite field) whose underlying set is amorphous, namely it cannot be partitioned into two infinite subsets.

This vector space is not finitely generated, but every proper subspace has a finite dimension.

Pick any such subspace $V'$ and consider $W=V/V'$. Then it is not very difficult to show that the following holds:

1. $W$ is a vector space whose underlying set is amorphous, and every proper subspace is finitely generated. But $|W|\neq|V|$.
2. There is no non-zero homomorphism from $W$ to $V$.
3. The natural quotient map is surjective from $V$ to $W$.

So $W$ is not projective. This also gives an example of an exact sequence which doesn't split.

The answer is no. Or at least not necessarily. At least for projectivity.

It is consistent that there is a vector space $V$ over $\Bbb F_2$ (or really any finite field) whose underlying set is amorphous, namely it cannot be partitioned into two infinite subsets.

This vector space is not finitely generated, but every proper subspace has a finite dimension.

Pick any such subspace $V'$ and consider $W=V/V'$. Then it is not very difficult to show that the following holds:

1. $W$ is a vector space whose underlying set is amorphous, and every proper subspace is finitely generated. But $|W|\neq|V|$.
2. There is no non-zero homomorphism from $W$ to $V$.
3. The natural quotient map is surjective from $V$ to $W$.

So $W$ is not projective. This also gives an example of an exact sequence which doesn't split. As Jeremy Rickard observes in the comment, this is also a counterexample to injectivity.

(The above can be done easily over infinite fields too, but then $V$ is not amorphous, just weird, and we might not have $|W|\neq|V|$ in general.

Another generalization is that we might allow subspaces to be not just finitely generated, but generated by a well-ordered set up to some cardinal $\lambda$ (which would mean the subspace is free, of course), but the space itself is not generated by any such set.)

I think that a relevant paper to mention here would be the following:

Andreas Blass, Injectivity, projectivity, and the axiom of choice, Trans. Amer. Math. Soc. 255 (1979), 31--59.