Let $M$ be a smooth manifold. Let $(T^*M, \pi, M)$ be cotangent bundle, where
$$ T^*M := \bigsqcup_{m\in M} T^*_m M$$ is a set equipped with corresponding topology $O_{T^*M}$ and smooth structure $A_{T^*M}$.
Then at each point $m\in M$ we can define $k$-exterior power of the cotangent space $\Lambda^k(T^*_mM)$.
Then we can define: $$ \Lambda^k(T^*M) := \bigsqcup_{m\in M} \Lambda^k(T^*_mM) $$ Question: How do I equip this set with a topology and smooth structure?
I want to be able to construct bundle $(\Lambda^k(T^*M), \pi', M)$ and then define module of smooth $k$-forms as a module of smooth sections $\Gamma(M, \Lambda^k(T^*M))$, but to do so I need first to transform the set above into a smooth manifold and I don't know how to do it properly. Any tips?
Edit 1: We can define initial topology on $\Lambda^k(T^*M)$ with respect to projection map $\pi'$: $$ \pi': \Lambda^k(T^*M) \to M \\ (m, \alpha) \mapsto m $$ as such, we can pick
$$ B := \{ \text{preim}_{\pi'}(U) \in 2^{\Lambda^k(T^*M)} | \forall U \in O_M\}$$
to be a subbasis. Then we can define topology $O_{\Lambda^k(T^*M)}$ as generated by this subbasis, i.e. set of all unions of finite intersections of elements of $B$.
That is $(\Lambda^k(T^*M), O_{\Lambda^k(T^*M)})$ - topological space, such that $\pi'$ is continious.
Now the only thing left is to find smooth atlas $A_{\Lambda^k(T^*M)}$ with respect to which $\pi'$ will become smooth map.