In Hatcher page 115: Elements of $H_n(X,A)$ are represented by $n$-chains $\alpha \in C_n(X)$ such that $\partial \alpha \in C_{n-1}(A) \subset C_{n-1}(X)$. We think of an element of $H_n(X,A)$ as being an $n$-thing in $X$ whose boundary (an $(n-1)$-thing) lies in $A$.
Since the relative homology groups fit into the long exact sequence $$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to H_{n-1}(X) \to H_{n-1}(X,A) \to \cdots,$$
My questons:
Are these maps surjectiveness or injectiveness, or non-surjectiveness or non-injectiveness? We can clarify one by one?
How to be precise about the maps given above? I meant there are precise relations of $$ker f_{j+1} = im f_{j} $$ for every map? even the dimensionality jumps?
Yes, at each object in a long exact sequence the kernel of the map on the right is equal to the image of the map on the left. This is the definition of exactness.
None of the involved maps have to be injective or surjective a priori. However, the definition of exactness immediately shows that a map $f\colon Z\to W$ in an exact sequence …
This is because injectivity is equivalent to $\operatorname{ker}(f)=0$ which then means that the map on the left has to have $0$ as its image, so it's the zero map. Surjectivity is equivalent to $\operatorname{im}(f)=W$ which means that the map on the right has to have $W$ as its kernel, so it's the zero map.
Note that for a map $P\xrightarrow{g} Q$ in the sequence it is sufficient that $P=0$ or $Q=0$ to have $g=0$. So when an object is $0$, the map two to the right is injective and the map two to the left is surjective: $$ \cdots\xrightarrow{\text{surj.}} * \xrightarrow{0} 0 \xrightarrow{0} * \xrightarrow{\text{inj.}} \cdots $$