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I am reading Milnor's book Morse Theory on p.67 he defines tangent space.

"By the tangent space of $\Omega$ (which is the path space) at a path $\omega$ will be meant the vector space consisting of all piecewise smooth vector fields $W$ along $\omega$ for which $W(0) = 0$ and $W(1) = 0$."

What I don't understand is the last part of the definition $W(0) = 0$ and $W(1) = 0$. What does it mean for the vector field $W$ to act this way? Is this the same thing as when people refer to a vector field vanishing? If so, what does that mean?

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I don't have the book in front of me, but I believe the vector field is along the curve. So at every point of the curve you pick a vector in the manifold in which the curve lives. The curve is a map $\omega:[0,1]\rightarrow M$ into the manifold $M$ and the vector field is chosen so that for every $t\in [0,1]$ you pick a vector $W(t)\in T_{\omega(t)}M$. Another way of saying this is that $W$ is a section of the pullback bundle $\omega^*(TM)$

Why should the tangent space along a path be such a vector field? Imagine a family of curves $\omega_s$ with $\omega_0=\omega$. Then the infinitesimal change is given by $W(t)=\frac{d\omega_s(t)}{ds}$. If the endpoints are fixed. i.e. $\omega_s(0)=\omega(0)$ and $\omega_s(1)=\omega(1)$ we can conclude that $W(0)=0$ and $W(1)=1$.

You really should draw some families of paths and try to draw the vector fields I just described to get an intuition.

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    That was great, thank you. I don't quite understand the last sentence. I don't think I understand what it means when they say when a vector equals 0. What does that mean? What does that look like? Like W(0) = 0. W(0) is a vector on the curve at t=0, but what does it mean when it says that it equals 0?2017-02-11
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    Try to draw the family of curves $\omega_s:[-1,1]\rightarrow \mathbb{R}^2$ defined by $\omega_s(t)=(t,s (t^2-1))$. Then compute the derivative $W(t)=\frac{d\gamma_s(t)}{ds}\bigr|_{s=0}$ and draw this as well. Then $W(-1)=W(1)=0$.2017-02-12
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    @user394412: It is just the zero in the tangent space to $\omega(0)$.2017-02-19