Let X be a topological space, and $P \in X$ a point. I had to prove $\tilde H_{i}(X) \simeq H_{i}(X,P)$. Instead of working on the complex i tried to use some fact of group theory but I'm not so sure of the validity of my conclusion. Is this argument ok or too "obscure" as I'm using some canonical isomorphism?
Let's consider the following continuous natural map: $i:P \to X$ and $r:X \to P$. Their functorial homomorphism induce the following split: $H_{i}(X)\simeq ker(r_{*})\bigoplus im(i_{*})\simeq \tilde H_{i}(X)\bigoplus H_{i}(P)$ as that is our definition of reduced homology and as the image of an injective group homomorphism it's isomorphic to the group itself.
We consider now the following long exact sequence derived from relative homology:
For $i>0$ we have that $H_{i}(\{P\})=0$ and the isomorphism is granted. Consider now the end of the sequence:
As $Im(j_{*})=ker(0)$ we know that $j_{*}$ has to be a surjective homomorphism. And we have that $ker(j_{*})\simeq im(i_{*}) \simeq H_{i}(\{P\})$. The isomorphism $\tilde H_{0}(X)\simeq H_{0}(X,P)$ follows now from the first isomorphism theorem.


$$ H_0({P}) \stackrel{i_*}{\to} \tilde{H}_0(X)\oplus ... $$– William Apr 02 '20 at 13:54