Let $E\to X$ be a complex line bundle. Let $\tilde{H}^2(X,\mathbb{Z})$ denote the image of $H^2(X, \mathbb{Z})$ in $H^2(X, \mathbb{R})$ under the natural homomorphism induced by the inclusion of the constant sheaves $Z\subset R $
I am reading the proof of $c_1(E)\in \tilde{H^2}(X,\mathbb{Z})$ from Well's book, and on Well's book, it says that:
I wonder why it suffices to show that $c_1(U_{1,N})\in H^2(P_{N-1},\mathbb{Z})$?
By $c_1(U_{1,N})\in H^2(U_{1,N},\mathbb{R})$ and $\Phi:X\to G_{1,N}=P_{N-1}$, we see $\Phi^*(c_1(U_{1,N})\in H^2(X,\mathbb{Z})$ if $c_1(U_{1,N})\in H^2(P_{N-1},\mathbb{Z})$, but then why do we have $\Phi^*(c_1(U_{1,N}))\in \tilde{H^2}(X,\mathbb{Z})$? I don't think it's true that ${H^2}(X,\mathbb{Z})\subset \tilde{H^2}(X,\mathbb{Z})$.
The definition of $\Phi$ for the case of bundle with rank $r$ is given below:
It is not even meaningful to say that $c_1(U_{1,N})\in H^2(P_{N-1},\mathbb{Z})$. By definition, $c_1(U_{1,N})$ is an element of $H^2(P_{N-1},\mathbb{R})$, and $H^2(P_{N-1},\mathbb{Z})$ is not a subset of $H^2(P_{N-1},\mathbb{R})$; only $\tilde{H}^2(P_{N-1},\mathbb{Z})$ is. So, I would assume this is either a typo or a mild abuse of notation: the natural map from $H^2(P_{N-1},\mathbb{Z})$ to $\tilde{H}^2(P_{N-1},\mathbb{Z})$ is an isomorphism, so they can be treated as interchangeable.