Existence criterion of $\operatorname{Spin}_{\mathbb{C}}$ structure via determinant line bundle In Dan Freed's notes Exercise 9.30 he outlines the proof of the existence criterion which is that there exists $\tilde{c} \in H^2(M;\mathbb{Z})$ such that $2\tilde{c} = c_1(E)$.  His approach is to pass to the determinant line bundle Det$(E) \to M$ of a complex vector bundle and work on $c_1(\operatorname{Det}(E))$ via its equality to $c_1(E)$.  I actually do not appreciate the benefit of passing to determinant line bundles and in particular I have no idea about (1) how to prove the first Chern class (step ii) there and (2) how to define the suggested Lie group homomorphism (step iii) there.  Could somebody please help me?
By the way I have asked a question on the same topic but for a different approach given in Cohen's notes on the topology of the fiber bundles.  If you have some idea about it please do me a favor drop it there!
 A: Here are several hints.  
For step (ii), the splitting principle allows you to prove the identity $c_1(E) = c_1(\operatorname{Det} E)$ under the additional assumption that $E = L_1 \oplus \cdots \oplus L_n$.  It might be useful to know that $$\operatorname{Det}(L_1 \oplus \cdots \oplus L_n) \cong L_1 \otimes \cdots \otimes L_n.$$  This follows from induction and the fact that $\Lambda^n (V \oplus W) \cong \bigoplus_{i + j = n} \Lambda^i(V) \otimes \Lambda^j(W)$.  
For step (iii), the existence of a continuous map $\tilde{U}(1) \to \mathbb{T}$ follows from the lifting criterion for covering spaces.  However, to show that you have a Lie group homomorphism requires a little bit more work.  First, you need to pick to the lift that sends the identity to the identity.  Once you do, the uniqueness of lifts will show that it is a group homomorphism.  Smoothness can checked locally, and all the covering maps are local diffeomorphisms while $\det$ is polynomial and hence smooth.  So indeed we have a homomorphism of Lie groups.   
Finally, the reason why you want to consider the determinant bundle is that it's a line bundle, and for line bundles $L_1, L_2$ we have the formula $c_1(L_1 \otimes L_2) = c_1(L_1) + c_1(L_2)$.  The argument in step (iv) is to show that using (iii), one can produce a line bundle $\sqrt{\operatorname{Det} E}$ whose tensor square is $\operatorname{Det}(E)$.  Using the formula for the first Chern class of tensor products of line products, this completes the proof of the proposition.  
