Suppose that $T$ is a bounded finite-rank linear operator from a normed space X into a normed spaceY. Prove that $X/ker(T) \cong T(X)$, whether or not either X or Y is complete.
$T$ is a bounded so $T$ is continuous then $ker(T)$ is close. Also $T$ is linear thus $T(X)$ is finte dimensional but every finite dimensional linear space is complete.
You just write the isomorphism explicitly: $\tilde T:X/\ker T\to T(X)$, given by $$ \tilde T: x+\ker T\longmapsto Tx. $$ This is obviously linear. If $\tilde T(x+\ker T)=0$, then $Tx=0$ so $x\in \ker T$; that is, $x+\ker T=\ker T$; so $\tilde T$ is injective. Surjectivity is clear.