Showing that the diagonal of $X\times X$ is transversal to the graph of $f$. (1.5.10 Guillemin and Pollack) The question and its answer is given below:

 

But I am wondering, is it also correct if I showed that graph f is transversal to diagonal of $X\times X$?
Also, I can not understand the general idea he using in his proof in the last paragraph, could anyone explain this for me please?
EDIT:
These are the exercises referred to: 




Thank you!   
 A: I guess this comes a bit late, but still...
This is a potential answer for the 1.5.9 from Guillemin & Pollack's Differential Topology
1.5.9: Suppose that $+1$ is eigenvalue of A
For $W$ to be transversal to $\triangle$ we would need
\begin{equation}
\forall (x,x) \in W\cap\triangle : T_{(x,x)}W + T_{(x,x)}\triangle = T_{(x,x)}{(V\times V)}
\end{equation}
Now $T_{(x,x)}\triangle$ is the diagonal of $T_xV\times T_xV$ (cf. the 1.2.10 from Guillemin & Pollack),
V is a vector space, so $T_xV = V$ and hence $T_{(x,x)}\triangle = \triangle$
Also $T_{(x,A(x))}W $ is the graph of $d_xA : T_xV \rightarrow T_xV$ (cf. the 1.2.11 from Guillemin & Pollack). A is linear, so $\forall x \in V : d_xA = A$. Moreover, if $x \in W \cap \triangle$ then $A(x) = x$ and hence $T_{(x,x)}W = T_{(x,A(x))}W = W$. So the condition for $W$ and $\triangle$ to be transversal becomes
\begin{equation}
W + \triangle = V\times V
\end{equation}
But it is clear that $V\times V$ contains points that $W + \triangle$ does not. For instance, if $v \in V$ is a non-zero vector such that $Av = v$ then $(v,0)$ is not included in $W + \triangle$.
Now suppose that $+1$ is not an eigenvalue of $A$. Then $W \cap \triangle = \{ (v,v) : $ for $v \in V$ such that $Av = v \} =  \{ (0,0) \} $ . Then $\dim(W+\triangle) = \dim(W) +  \dim(\triangle) - \dim(W \cap \triangle) = \dim(V \times V)$ so $W + \triangle = V \times V$
So "$+1$ not eigenvalue of $A$" implies $W \pitchfork \triangle $
And hence $W \pitchfork \triangle $ if and only if $+1$ is not an eigenvalue of $A$
