Degree of a map globally constant

Question

I'm having trouble with the following proof: Denote by $deg_pf$ the deegree of a proper differentiable map $f:X\rightarrow Y$ between manifolds, where $p$ is a regular value of $f$ and the degree is the algebraic sum of "orientation numbers" of $df$. Suppose also $Y$ is connected.
Also, assume that for each regular value $p$, there exists a neighbourhood $U_p$ of $p$ where, for all $q\in U_p$ regular value of $f$, $deg_qf=deg_pf$. Then the argument to show that the degree is equal at all regular values is as follows:
For every $y\in Y$, choose $U_y$ neighbourhood of $y$ where $deg_qf=deg_pf$, $\forall p,q$ regular values in $U_y$. Then since the set of regular values is dense, each $U_y$ has a corresponding regular value associated to it. Then the proof procceds in using the fact that $Y = \bigcup U_y$ and $Y$ connected to show that all degrees are equal. My problem is with the existence of the neighbourhood $U_y$ for $y$ not a regular value of $f$. I can't justify its existence... Any ideas?