Suppose you have a diffeomorphism
$$ \phi: M \to N. $$
As far as I understand, this gives rise to two distinct isomorphisms
$$ a : \mathcal A^1(M) \cong \mathcal A^1(N) :b $$
between the space of 1-forms on $M$, and on $N$. The two are given as follows:
Given a 1-form $\theta: M \to T^*M$, $a(\theta): N \to T^*N$ is defined to be the 1-form fitting into the commutative diagram below: $$ \require{AMScd} \begin{CD} T^*M @>T^*\phi>\cong> T^*N \\ @A{\theta}AA @A{a(\theta)}AA\\ c @<\phi^{-1}<\cong< d \end{CD} $$
On the other hand, $b(\theta)$ is defined to be the usual pull-back of 1-forms, so that for any tangent vector $v \in TM$, we have $$ b(\theta)(v) = \theta(D\phi(v)) $$ where $D\phi: TM \to TN$ is the derivative of $\phi$.
Is there any relationship between these two isomorphisms? For instance, an equation in which both isomorphisms appear?
Dually, we could note that there are two isomorphisms $$ c, e: \Gamma(TM) \to \Gamma(TN) $$ where given any vector field $X: M \to TM$, $c(X)$ is defined to fit into the commutative diagram
$$ \require{AMScd} \begin{CD} TM @>T\phi>\cong> TN \\ @A{X}AA @A{c(X)}AA\\ c @<\phi^{-1}<\cong< d \end{CD} $$
and $e(X)$ is the usual pushforward of a vector field along its image: For any smooth function $f: N \to \mathbb R$, we have $$ e(X)(f) = X(f \circ \phi). $$
Is there any relationship between $c$ and $e$?