Let $T : V \to U$ be a linear transformation and $A,B,C \subseteq V$... 
a-) Suppose that $A$ is linearly independent on $V.$ What conditions in $T$ guarantee that $T(A)$ is linearly independent on $U$?.


b-) Suppose $C$ generates $V.$ What conditions in $T$ guarantee that $T(C)$ generates $U$?


c-) Suppose $B$ is a basis of $V.$ What conditions in $T$ guarantee that $T(B)$ is a base of $\operatorname{Im}(T)?$


d-) Suppose $B$ is a basis of $V.$ What conditions in $T$ guarantee that $T(B)$ is a $U$ base?

a-) $T$ does not guarantee that $A,$ being LI in $V,$ will be LI in $U.$ To guarantee this we need that all elements of $A$ lead to a single element of $U,$ and also that every element of $U$ has its representative in $A.$ Therefore: The statement is true if and only if the transformation is bijective.
b-)The elements of $C$ need to be able to build all the elements of $U.$ Therefore: The statement is true if and only if the transformation is surjective.
c-)As $\operatorname{im}(T)$ already implies surjectivity, we need $T(B)$ to be injective. Therefore: The statement is true if and only if the transformation is injective.
d-) As the base needs to be LI and generate all the space by items a-) and b-) we have that: The statement is true if and only if the transformation is bijective.
The idea of the answers is correct?
I think at least the concepts are correct (I could be wrong). But my main difficulty is in the formalization of ideas.
Thanks for the attention.
 A: It seems to me you are approaching this problem the wrong way.  It is not particularly helpful to make guesses about the answers based on the kind of vague reasoning that you are using.  It would be better to try to write proofs, and see what is needed to make the proof work.
For example, for a), try to prove that $T(A)$ is linearly independent.  The definition of linearly independent tells you how to get started on that: assume you have a linear combination of elements of $T(A)$ that is equal to 0, and try to prove that all the coefficients are 0.  Write out such a linear combination explicitly, set it equal to 0, and try to use the fact that $A$ is linearly independent and $T$ is linear to prove that the coefficients are 0.  You should be able to get partway through the proof, but at a certain point you will be stuck.  The question is: what property of $T$ would allow you to get unstuck?
A: a) Since $A$ may well not be a subset of $U$, it makes no sense to say that $A$ is (or is not) linearly independent in $U$. The condition that guarantees that $T(A)$ is linearly independent is that $T$ is injective.
b) Again, since $C$ may well not be a subset of $U$, it makes no sense to say that you can use $C$ to build $U$. But, yes, the statement is true if and only if $T$ is surjective.
c) The assertion “As $\operatorname{Im}(T)$ already implies surjectivity” is meaningless, just as the assertion that $T(B)$ is injective. But, yes, the condition that guarantees that $T(B)$ is a basis of $\operatorname{Im}(T)$ is that $T$ is injective.
d) Yes, it happens if and only if $T$ is bijective.
A: a) $T:V\to W$ is injective if and only if $T$ leads linearly independent in linearly independent.

$(\Rightarrow)$ Let $v_j\in A$ (linearly independent). If $\sum \lambda_jT(v_j)=0$, with $\lambda_j\in\mathbb K$, then $T(\sum\lambda_j v_j)=0$ since $T$ is injective therefore $\sum\lambda_j v_j=0$. Thus $\lambda_j=0$.


$(\Leftarrow)$ Let $v\in A$ such that $T(v)=0$. If $v\neq 0$ then since $\{v\}$ is linearly independent and $\{T(v)\}=\{0\}$ is linearly dependent, which is a contradiction.

b) $T:V\to W$ is surjective if and only if $T$ leads generates in generates.

$(\Rightarrow)$ Let's suppose $C$ generates $V$. Let $u\in U$, since $T$ is surjective then $u=T(v)$ for some $v\in V=Span(C)$. Thus $\exists_nv_j\in C$, $\lambda_j\in\mathbb K$ such that $u=T(\sum\lambda_jv_j)=\sum\lambda_jT(v_j)\in Span(T(C))$. Therefore $U=Span(T(C))$.


$(\Leftarrow)$ Let $u\in U=Span (T(C))$, then $u=\sum\lambda_jT(v_j)$, for some $\lambda_j\in\mathbb K, v_j\in C$. Then $u=T(\sum\lambda_jv_j)$, where $\sum\lambda_jv_j\in Span(C)\subset V$. Therefore $T$ is surjective.

c),d) With a) and b), this is solved for $T$ bijective.
