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The image of a linear map $T:V\to W$ is the vector subspace $\mathrm{im}T=\{T(v):v\in V\}$ of its codomain.

The continuity condition at $x\in X$ between metric spaces isThe function is continuous when this holds at every $x$.

If $f$ is continuous and $U\subseteq Y$ is an open set, take $x\in f^{-1}(U)$. Some $ϵ$-ball around $f(x)$ lies in $U$. Continuity gives a $\delta$-ball around $x$ mapped into that ball and hence contained in $f^{-1}(U)$. Thus $f^{-1}(U)$ is open, including the empty set, in the notation for the image and preimage of a function.

Conversely, suppose every open set has open preimage. For fixed $x$ and $ϵ>0$, the preimage of the open ball $B_e(f(x),ϵ)$ is open and contains $x$, so contains some $B_d(x,\delta )$. This gives the displayed metric definition. Therefore $\boxed{f\text{ is continuous}⟺f^{-1}(U)\text{ is open for every open }U}$.

A linear map $T:V\to W$ is surjective when its image is all of $W$. For finite-dimensional spaces this is equivalent to $\mathrm{rank}T=\dim W$.