Show that $\ker A\subset Im (I-A)$ and that $Im(A)\subset \ker (I-A).$ Problem: Show that $\ker A\subset Im (I-A)$ and that $Im(A)\subset \ker (I-A)$, where $A$ is a linear transformation on a vector space. 
My attempt: Let $x\in \ker A,$ then $Ax=0.$ But since $x\in V$ (the vector space), it's image obtained by the map $(I-A)(x)$ must be in $Im(I-A).$ But $(I-A)(x)=Ix-Ax=x$ and so $x\in Im(I-A).$ Thus we have shown the first part to be true. 
Now consider an element $y\in Im(A).$ Then for some $x\in V$ we have that $y=Ax.$ This is the part where I am having trouble to carry on. It is not necessarily true that $x=Ax$, which would automatically imply that $x\in \ker(I-A).$ So I am not sure whether this holds. Any hints on proving this fact would be much appreciated. 
Here is the exact problem (3.77): 

 A: Suppose $\operatorname{Im}\mathcal{A}\subseteq\operatorname{Ker}(\mathcal{I}-\mathcal{A})$. Take $v\in V$; then $\mathcal{A}(v)\in\operatorname{Im}\mathcal{A}$, so, by assumption,
$$
(\mathcal{I}-\mathcal{A})(\mathcal{A}(v))=0
$$
that is,
$$
\mathcal{A}(v)-\mathcal{A}^2(v)=0
$$
which implies $\mathcal{A}=\mathcal{A}^2$.
Thus the condition $\mathcal{A}=\mathcal{A}^2$ is necessary in order that $\operatorname{Im}\mathcal{A}\subseteq\operatorname{Ker}(\mathcal{I}-\mathcal{A})$. It is also sufficient, as you can easily prove.
Therefore exercise 3.77 is flawed.
Note that if the two statements were to hold for every linear map, changing $\mathcal{A}$ into $\mathcal{I}-\mathcal{A}$ in the first statement we'd get that $\operatorname{Im}\mathcal{A}=\operatorname{Ker}(\mathcal{I}-\mathcal{A})$, for every linear map on the vector space, which is patently false. When $\mathcal{A}$ doesn't have $1$ as eigenvalue, but is not injective, $\mathcal{I}-\mathcal{A}$ is injective. In the finite dimensional case, this is of course a contradiction, unless $\mathcal{A}$ is the zero map.
