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To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false if

1. you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false: in Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example of a torus with a nonzero integral $(1,1)$ class which is not a divisor.

To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false if

1. you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false: In Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example of a torus with a nonzero integral $(1,1)$ class which is not a divisor, but it would necessarily lie in the image of $c_1$.

To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false

1. if you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifolds because there may not be enough vector bundles. (To be clear, I mean that conclusion in cohomology is false. For $X$ take a general torus.)

To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false if

1. you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false: in Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example of a torus with a nonzero integral $(1,1)$ class which is not a divisor.

To expand slightly on slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong Ch(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $Ch(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false

1. if you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifolds because there may not be enough vector bundles.

To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false

1. if you omit the $\mathbb{Q}$, or
2. if you work on a general (compact) complex manifolds because there may not be enough vector bundles. (To be clear, I mean that conclusion in cohomology is false. For $X$ take a general torus.)