Endomorphisms and K-Vector spaces I want to proof (i) and (ii)
(i) Let $V$ be a finite-dimensional K-Vector Space and $f\in End(V)=Hom(V,V)$. If there is a $g\in End(V)$ with $f\circ g=Id_V$, then $f$ is an Isomorphism with $g=f^{-1}$
(ii) Statement (i) is not true for an infinite-dimensional K-Vectorspace $V=Abb(\mathbb{N},K)$ with $Abb(\mathbb{N},K):=\{f|f:\mathbb{N}\rightarrow K\}$.
My ideas:
(i) Let's assume that $\exists g \in End(V): f\circ g=Id_V$. If $f\circ g=Id_V$ then $g\circ f=Id_V$ because $f: V\rightarrow V$. So $g$ is the inverse function of $f$ and $f$ is an automorphism, as it is an isomorphism and an endomorphism.
I am not entirely sure about this proof.
(ii) Here I need help, I know that if $K=\mathbb{R}$ then there cannot be a surjective function $f$ but what if $K=\mathbb{N}$?
 A: I do not think that your proof for (i) is complete:
First observe that $f \circ g = id_V$ implies that $f$ is surjective: Indeed, any $v \in V$ lies in the image of $f$ as $v = f(w)$ for $w := f(v) \in V$. (Similarly, one can deduce from $f \circ g = id_V$ that $g$ is injective, but this fact is not necessary to prove (i).)
However, it is well-known that surjective $K$-linear maps between finite dimensional vector spaces are necessarily bijective, hence isomorphisms. (Actually, this is not true for infinite dimensional vector spaces, and your task in (ii) is to provide a counterexample.) As a consequence, there exists a $K$-linear map $f^{-1} \colon V \to V$ such that $f \circ f^{-1} = f^{-1} \circ f = id_V$. But from this, we may easily conclude that
$$f^{-1} = f^{-1} \circ id_V = f^{-1} \circ (f \circ g) = (f^{-1} \circ f) \circ g = id_V \circ g = g,$$
proving that $g$ is the inverse of $f$, completing the proof of (i).
Concerning (ii), you may consider the example provided by Severin Schranven in his comment, i.e., the left shift $(a_0,a_1,a_2,...) \mapsto (a_1,a_2,...)$. Note here that one may regard any function $f \colon \mathbb{N} = \{0,1,2,3,...\} \to K$ as a sequence $(f(0),f(1),f(2),f(3),...)$.
If the left shift was an isomorphism, then it would in particular be (surjective and) injective. Can you see why the left shift fails to be injective?
And concerning the map $g$, you may think about how one could define a right shift.
P.S.: Note that it is impossible to choose $K = \mathbb{N}$, because (when speaking about $K$-vector spaces) $K$ is always assumed to be a field.
