Question on finite Vector Spaces, injective, surjective and if $V$ is not finite

The statements (i) and (ii) illustrate that finite-dimensional vector spaces behave in a way similar to finite sets, for a map $X \to X$ on a finite set is injective iff it is surjective. However, if $X$ is an infinite set there are maps that are injective but not surjective (e.g. $\mathbb N \to \mathbb N, x \mapsto x+1$) and vice versa (e.g. $\mathbb N \to \mathbb N, x \mapsto \max\{1,x-1\}$). You can turn this into a counterexample for the corresponding statement for vector spaces: let $V$ be the vector space freely generated by $\mathbb N$, i.e. the vector space of sequences $(a_1,a_2,\ldots) \in k^{\mathbb N}$ where $a_i =0$ for all but finitely many $i$ and consider the linear maps induced by the above maps, i.e. $$(a_1,a_2,\ldots) \mapsto (0,a_1,a_2,\ldots)$$ and $$(a_1,a_2,\ldots) \mapsto (a_1+a_2,a_3,a_4\ldots)$$ respectively. The first map is injective but not surjective, and vice versa for the second map.

The second one does not fail because the axiom of choise, you used the fact the the dimension of the space is finite.

A counter example: $V$ is the space of the the polynomials with coefficients in $\mathbb{R}$ and $T$ is defined by $$T(x^{i})=x^{i+1}$$ (that is $1\mapsto x,x\mapsto x^{2}$ etc' ).

In this case $T$ is clearly $1-1$ but not onto since $1$ is not in the image of $T$

Note $1$: We have that $$dim(Im(T))+1=dim(V)$$ but still $dim(Im(T))=dim(V)$ because both are not not finite, this is what fails.

Note $2$: we defined $T$ on a basis for $V$ so it's well defined.