In Section 26.2 of Tu's Introduction to Manifolds he computes the cohomology of the circle. I understand the proof of why $H^1(S^1) \cong \mathbb{R}$, but I am having some difficulty understanding his derivation of an explicit generator for $H^1(S^1)$.
He lets $U$ and $V$ be an open cover for $S^1$. Specifically both are arcs, thus their intersection has two connected components, which he labels as $A$ and $B$. He shows that a nonzero element of $H^1(S^1)$ is the image under the connecting homomorphism, $d^*$, of an element $(a,b) \in H^0(U \cap V) \cong \mathbb{R}^2$ with $a \neq b$. He then says
So we may start with $(a,b) = (1,0) \in H^0(U \cap V)$. This corresponds to a function $f$ with value $1$ on $A$ and $0$ on $B$. Let $\{\rho_U, \rho_V\}$ be a partition of unity subordinate to the open cover $\{U, V\}$, and let $f_U, f_V$ be the extensions by zero of $\rho_Vf, \rho_Uf$ from $U \cap V$ to $U$ and $V$, respectively. By the proof of Proposition 26.2, $j(-f_U, f_V) = f$ on $U \cap V$. From Section 25.3, $d^*(1,0)$ is represented by a 1-form on $S^1$ whose restriction to $U$ is $-df_U$ and whose restriction to $V$ is $df_V$. Now $f_V$ is the function on $V$ that is $\rho_U$ on $A$ and $0$ on $V - A$, so $df_V$ is a 1-form on $V$ whose support is contained entirely in $A$. A similar analysis shows that $-df_U$ restricts to the same 1-form on $A$, because $\rho_U + \rho_V = 1$. The extension of either $df_V$ or $-df_U$ by zero to a 1-form on $S^1$ represents a generator of $H^1(S^1)$. It is a bump 1-form on $S^1$ supported in $A$.
The first half of seems to follow from the zig-zag diagram used to construct the connecting homomorphism. It is the second part that I am having trouble with (in particular, the parts I have put in bold). I do not see how the two 1-forms restrict to the same 1-form on $A$. On $A$, $f_V$ is $\rho_U$, whereas $-f_U$ is $\rho_V$. So how do these two restrict to the same 1-form unless $\rho_U = \rho_V$?
This is not correct. Looking at how to extend a function in the previous section (around exercise $26.3$), we have specifically that on $A$, $f_U = \rho_V$ and $f_V = \rho_U$.
So, we have on $A$ $$f_U + f_V = \rho_V + \rho_U = 1$$ So, we get $d(f_U + f_V) = 0$ i.e. $df_U = - df_V$ (again, only on $A$).