if a smooth map between manifolds is immersion then the diagram is commutative 
This is a theorem from Bredon's Topology and Geometry.

And here is the proof given in the book.
Here we have to find charts $\phi$ and $\psi$ such that $\theta\circ \phi^{-1}=\psi^{-1}\circ \theta^{'}$.
In this proof all are fine before the last it is written that "the new charts satisfy the conclusion of the theorem". I did not find how the conclusion(i.e. the diagram is commutative) is satisfied. Please help me to understand this.Thanks.
 A: $$\require{AMScd}$$
\begin{CD}
p \in M^{m} @>\theta>> N^{n}\ni \theta(p)  \\
@V\Phi VV @V\Psi VV \\
\Bbb R^{m} @> \theta^{\prime}>> \Bbb R^{n} \\
@V i VV @A \zeta AA \\
\Bbb R^{n}   @>id>> \Bbb R^{m} \times \Bbb R^{n-m} 
\end{CD}
in the above diagram. $\theta^{\prime}$ is defined as follows :
$$  \theta^{\prime}: \Phi(U \cap \theta^{-1}(V)) \to \Psi(V) \\ 
     \theta^{\prime}(q) = \Psi \circ \theta \circ \Phi^{-1}(q)   
$$
now the $\zeta$ as function is defined on the open set $ i(\Phi(U \cap \theta^{-1}(V)) \times \Bbb R^{n-m} \to \Bbb R^{n}$ as follows :
$$ \zeta(x,y) = \theta^{\prime}(x) + (0,y)$$. it is clear that $\zeta(0,0) = (0,0)$ and $\zeta \circ id \circ i (x) = \theta^{\prime}(x)$.  Also on this point $D\zeta\vert_{(0,0)} = \begin{bmatrix} D\theta^{\prime}_0 && 0 \\ 0 && Id\end{bmatrix} $. So it is full rank. Now using  the inverse function theorem we find that $(0,0)$ on an open set of the form $E$ and $F$ around $(0,0)$ ,such that $\zeta$ is a diffeomorphism from $E$ to $F$.
Note that:
$$ i(x) = (x,0) \\
  id \circ i (x,0) = (x,0)\\
 \zeta \circ id \circ i (x,0) = \theta^{\prime}(x) +0 = \theta^{\prime}(x)
$$
Now if we define the new chart around $\theta(p)$ as follows :
$$ \Psi^{\prime} : \Psi^{-1}(F) \to \Bbb R^{n}$$
$$  \Psi^{\prime} (q) = \zeta^{-1} \circ \Psi (q).$$
and the new $\hat\theta$ as above. We would have :
$$ \hat{\theta}(x) = \zeta^{-1} \circ \zeta \circ id \circ i (x) = (x,0).$$
Where now, $\hat{\theta}$ is defined on the open set :
$$ \phi (U \cap \theta^{-1}(\Psi^{-1}(F))).$$
To better see the commutativity:
$$ {\Psi^{\prime}}^{-1} \circ \hat{\theta} \circ \Phi (x) =  
 \Psi^{-1} \circ \zeta \circ \zeta^{-1} \circ \theta^{\prime} \circ \Phi (x) = \theta(x) $$
