# construction of cotangent bundle

Cotangent Bundle $T^{*}(M)=\bigcup_{m\in M} M_m^{*}$ (disjoint union of contangent space) Could any one explain me how there is a natural projection from $T^{*}(M)\rightarrow M$ given by $\pi(f)=m$ if $f\in M^{*}_m$? I am not able to understand and feel the map naturally Please explain.

-

For each point in $M$ you have a cotangent space $M_m^*$. So we now look at the disjoint union of all of them as a collection of fibers -- "above" each point in $M$ lies a cotangent space.

The canonical map just "projects" each fiber (end every point within it) onto the point above which it lies, just the same as with the tangent bundle.

-