I am having much trouble trying to understand the idea of an exact sequence. As a toy example, I'm looking at the following exact sequence:
$$0 \xrightarrow{} ker(T) \xrightarrow{\iota} V \xrightarrow{T} U,$$
where $T$ is a homomorphism from a vector space $V$ to a vector space $U$, and $\iota$ is the natural inclusion map from the kernel of $T$ to $V$. I know this is a very vague request, but in short I was simply hoping for somebody to explain to me exactly what the heck is going on here. For some reason I am getting in my mind the thought that $ker(T) = 0$, meaning that $ker(\iota) = ran(\iota) = 0$, but I've been thinking about this so much to the point where the harder I think the more I find myself confused. Thanks in advance.
(Motivation: Besides simply trying to understand this stuff, I am also wanting to prove that given any homomorphism $S: W \xrightarrow{} V$ with $T \circ S = 0$, there exists a unique homomorphism $F: W \xrightarrow{} ker(T)$ such that $S = \iota \circ F$. Now, this seems to make sense intuitively, but I think I am having difficulty proving this because I am not using everything known about the exact sequence above.)
Let us consider a countable family $\mathcal F=\{A_i\}_{i=1,2,3,\dots}$ of vector spaces over a fixed ground field $\mathbb k$, let us say $\mathbb k=\mathbb R, \mathbb C$ for simplicity. With $0$ we denote the vector space with only one element, i.e. the unit of the abelian group composition (the "zero vector"). The following
$$A_1\stackrel{\alpha_1}{\rightarrow} A_2 \stackrel{\alpha_2}{\rightarrow} A_3 \stackrel{\alpha_3}{\rightarrow} \cdots$$
is called sequence of vector spaces, with $\alpha_i\in \operatorname{Hom}_\mathbb k(A_i,A_{i+1})$. The sequence is said to be exact if $$\operatorname{ker}(\alpha_i)=\operatorname{im}(\alpha_{i-1})$$ for all $i\geq 2$.
This condition is easy/hard to prove, given the above. The easier part is to show that $\alpha_{i+1}\circ\alpha_{i}=0$, i.e. $\operatorname{im}(\alpha_i) \subset \operatorname{ker}(\alpha_{i+1})$; to prove that $\operatorname{ker}(\alpha_{i+1}) \subset \operatorname{im}(\alpha_i) $ one needs to work on an existential condition on certain elements.
Take $\mathcal F=\{0,A_2,A_3,A_4,0\}$; the exact sequence $$0\rightarrow A_2 \stackrel{\alpha_2}{\rightarrow} A_3 \stackrel{\alpha_3}{\rightarrow} A_4 \rightarrow 0$$ is said to be a short exact sequence: by exactness we have $\operatorname{ker}(\alpha_2)=0$ and $\operatorname{im}(\alpha_3)=A_4$.
In other words, if the above sequence is short exact, then
$$0\rightarrow A_2 \stackrel{\alpha_2}{\rightarrow} A_3 $$ is exact, i.e. $\operatorname{ker}(\alpha_2)=0$, or $\alpha_2$ is a monomorphism (=it is injective). This sub-sequence is said to be left exact. Moreover $$ A_3 \stackrel{\alpha_3}{\rightarrow} A_4 \rightarrow 0$$ is exact as well: $\operatorname{im}(\alpha_3)=A_4$, i.e. $\alpha_3$ is an epimorphism (=it is surjective). This sub-sequence is said to be right exact.
Morally speaking, if a short exact sequence exists for $A_2$, $A_3$ and $A_4$, we can say that $A_3$ is "composed" in some way from $A_2$ and $A_4$, like in a twisted composition of those vector spaces. Split short exact sequences clarify the concept of twist, but I won't talk about them now.
Using the above definitions, let us consider the toy example
$$0 \xrightarrow{} ker(T) \xrightarrow{\iota} V \xrightarrow{T} U,$$
where $T$ is a homomorphism from a vector space $V$ to a vector space $U$, and $\iota$ is the natural inclusion map from the kernel of $T$ to $V$. This is an exact sequence; in fact, although $\operatorname{im}(\iota)=\operatorname{ker}(T)$ and $\operatorname{ker}(\iota)=0$ by construction.
Question in the OP: what does it happen if $T$ is injective, i.e. $\operatorname{ker}(T)=0$? If this is the case, then you can introduce the left exact sequence
$$0\rightarrow V \xrightarrow{T} U; $$
exactness is equivalent to injectivity of $T$. Similar considerations hold if $T$ is surjective. If $T$ is an isomorphism, then
$$0\rightarrow V \xrightarrow{T} U\rightarrow 0 $$
is a short exact sequence.