Homology of a disjoint union

I have a space $$Z=X\sqcup P$$, where $$P=\{*\}$$ is a disjoint point. I want to show that $$\tilde{H_i}(Z) \cong H_i(X)$$. I start with the following: Let $$f:A \rightarrow C$$ and $$g:B \rightarrow C$$ be homomorphism of abelian groups where we also assume $$g$$ to be isomorphism. We then also define $$f\oplus g:A\oplus B \rightarrow C$$ as $$(a,b) \mapsto f(a) + f(b)$$. We can then construct an isomorphism $$\omega$$ between $$A$$ and the kernel of $$f \oplus g$$ by sending $$a$$ to $$(a,g^{-1}(-f(a)))$$. One can define the $$n$$-th reduced homology in the following way: let $$X$$ be a space and let $$P = \{∗\}$$ be a one-point space. Then there is a unique continuous map $$\gamma^X:X \rightarrow P$$. The $$n$$-th reduced homology group of $$X$$ is: $$$$\tilde{H_i}(X)=ker(\gamma_{*}^X:H_n(X) \rightarrow H_n(P)).$$$$ So I argue as follows: $$$$\tilde{H_n}(Z)=ker(\gamma^Z:H_n(Z)\rightarrow H_n(P)) \cong ker(\gamma:H_n(X)\oplus H_n(P) \rightarrow H_n(P)).$$$$ Let $$f^{'}: H_n(X) \rightarrow H_{n}(P)$$ be a group homomorphism (is that always well-define, can I do that?) and $$g^{'}:H_n(P) \rightarrow H_n(P)$$ an isomorphism (so $$g^{'}$$ is just $$=$$). Then we get $$H_i(X) \cong ker(f^{'} \oplus g^{'}) \cong ker(\gamma)\cong \tilde{H_n}(Z)$$. Does this make sense at all?

• It is not clear to me what you are doing with these homomorphisms. Why are you not defining them? Perhaps you should use the fact there is a right inverse to the map $X \sqcup *\rightarrow *$ given by including the point as the disjoint point. Analyze what happens on homology. Commented Dec 20, 2019 at 2:23
• Thank @ConnorMalin. I am not sure I understand, I am still really new to this stuff and struggle to work things out from definitions. So, are you saying that $X \sqcup *$ are homeomorphic and so this induces the isomorphism between the homology groups of $X \sqcup *$ and $*$? And I should somehow go from there? Commented Dec 20, 2019 at 11:11