The rank is $r(\alpha )=\dim \mathrm{im}\alpha$ and the nullity is $n(\alpha )=\dim \ker \alpha$. Extending a basis of the kernel to a basis of $V$, the images of the added vectors form a basis of the image, proving the rank-nullity theorem

Since $\mathrm{im}(\alpha +\beta )\subseteq \mathrm{im}\alpha +\mathrm{im}\beta$, the upper sum bound follows. Applying it to $\alpha =(\alpha +\beta )-\beta$ and interchanging $\alpha ,\beta$ gives the lower bound. For products, the image lies in $\mathrm{im}\alpha$ and is the image under $\alpha$ of $\mathrm{im}\beta$, giving the upper bound; rank-nullity on that restriction gives

All four bounds are sharp. In a fixed basis, nested diagonal projections attain the product upper bound; placing $\mathrm{im}\beta$ with the smallest possible intersection with $\ker \alpha$ attains its lower bound. Taking $\beta =-\alpha$ on a subspace of dimension $\min (r(\alpha ),r(\beta ))$ attains the sum lower bound, while choosing the two images in general position and avoiding cancellation attains the sum upper bound.

Both upper bounds need not be attainable simultaneously over every field. Over ${\mathbb{F}}_2$ with $n=1$ and both ranks one, necessarily $\alpha =\beta =I$. Then $r(\alpha \beta )=1$ but $r(\alpha +\beta )=0$, below its upper bound one.

If such an $f$ existed, boundedness would make its isolated singularity at zero removable. Let $F$ be the holomorphic extension to the disc. Continuity gives $1\le |F(0)|\le 2$. Equality at either endpoint contradicts the minimum-modulus principle applied to $F$ or the maximum modulus principle, since $F$ is nonconstant. Hence $F(0)\in A$. Surjectivity of $f$ supplies some nonzero $z$ with $f(z)=F(0)$, contradicting injectivity of the extension. Therefore

The closure $\overline{Z}$ is the intersection of all closed subsets containing $Z$; equivalently, every neighbourhood of a point of $\overline{Z}$ meets $Z$. The subspace is dense when $\overline{Z}=X$. A space is Hausdorff when distinct points have disjoint neighbourhoods.

For Hausdorff $Y$, the diagonal ${\Delta }_Y$ is closed in $Y\times Y$. Thusis closed. Since it contains the dense set $Z$, it equals $X$.

The conclusion fails without Hausdorffness. Let $X=Y=\{0,1\}$ have the Sierpinski space topology $\{\varnothing ,\{1\},X\}$ and let $Z=\{1\}$, which is dense. The identity map and the constant map with value $1$ are continuous and agree on $Z$, but differ at $0$.