is the homomorphism induced by the inclusion map is the inclusion map? let $X$ be a topological space, $A\subset X$ subspace. 
Consider $i:\:A\to X$ the inclusion, and $i_* :\:H_n(A)\to H_n(X)$ the induced homomorphism.
is $i_*$ the natural homomorphism $[a]\mapsto [a]$?
the definition of the induced homomorphism between homology groups is intuitive but a little bit complex (as we define it in two parts), but I think it reduced to what I've just said. Am I right?
 A: I think your question is a little ambiguous so I will answer both potential questions.
1 An inclusion of spaces does not induce an inclusion of groups. As pointed out in the comments you can include a space with non-trivial homology into a space with trivial homology (i.e. same homology as a point). That would imply that there has to be an induced map in some dimension which is the trivial map 0, in particular not an inclusion, since not injective.
2 If you consider a homology class $[a]$ represented by a $n$-cycle $a$ in $A$, you can also consider it as a cycle in $X$. That being said, this gives you a homology class $[a] \in H_nA$ and also $[a]\in H_nX$. In that terms the homomorphism induced by the inclusion is $[a] \mapsto [a]$, which is certainly bad notation (better would be maybe to make clear which equivalence relation we are talking about e.g. $[a]_X \in H_nX$). So in that case the answer is yes, but however note that by 1 we have that the cycle $a$ can be non-trivial in $A$ but trivial (in terms of homology, i.e. null homologous) in $X$. This construction and definition of the map makes the homomorphism natural.
