Why is $(-\infty,1)$ the domain of the solution to $y'=y^2$, $y(0)=1$ instead of $(-\infty,1)\cup (1,\infty)$ Consider the ODE $$y'=y^2$$ with initial condition $$y(0)=1$$.
A solution is $$y(t)=\frac{-1}{t-1}$$
Which the text I'm reading mentions is only valid on $(-\infty,1)$, which includes the left branch of the graph of this hyperbola. They justify this by saying that the initial condition is included in the left-hand side of the graph of this function, so that we cannot extend the solution to the entire real line. My question is, what is wrong with saying that the solution is $y(t)$ on $(-\infty,1)\cup (1,\infty)$? Why can we not include the rest of the domain where the solution $y(t)$ is defined? Thanks
EDIT: They mention that this is the "maximum interval on which the solution curve is defined", perhaps by calling it a solution "curve" they are implicitly trying to say we want the greatest domain on which the solution is continuous?
 A: You certainly can say it is a solution. But you cannot say it is the solution. As Winther has pointed out in his comment above you lose uniqueness when going past the singularity. You could for example extend your solution by setting it identically zero to the right of the singularity, this would be a solution as well.
A: It depends, actually, on the "universal" domain you are working in.  With real variables only, as others  ( notably @Severin Schraven) have explained you cannot determine the solutions uniquely past the singularity $x=1$.  You are blocked.  But in complex variables you can propagate around the pole and define the solution everywhere around said pole.  Then the solution $f(z)=-1/(z-1)$, using nomenclature usual for complex analysis, has as domain all complex inputs except $1$.
A: As others have pointed out, consider the domain where the solution is Lipschitz. If a function is not globally Lipschitz, then your ODE won't have a unique solution. This comes straight from the Picard-Lindelof theorem. 
