I'd like to prove that $\tilde{H}_n(X \cup CA) \simeq H_n(X,A)$, where $CA$ denotes the cone of $A$, i.e $CA = A \times I / A \times \{0\}$ where $A$ is a subset of $X$ topological space.
The first step in order to do so is to prove the following isomorphism $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$. Since $CA$ is contractible I thought of taking the long exact sequence of the pair and i get $$\longrightarrow H_n(CA) \longrightarrow H_n(X \cup CA) \longrightarrow H_n(X \cup CA,CA) \longrightarrow H_{n-1}(CA)\longrightarrow$$
So if $n-1 \ne 0$ i.e $n > 1$ since $H_i(CA) = 0$ we have the $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$ since for those $n, \hspace{0.1cm} \tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA)$.
Here is where I got stuck, since the terms are no longer $0$ I don't know how to continue in the case $n=0,n=1$. Any help would be appreciated.
The long exact sequence has two variants: One for unreduced homology groups ands one for reduced homology groups. The latter is $$\dots \to \tilde H_n(A) \to \tilde H_n(X) \to H_n(X,A) \to \tilde H_{n-1}(A) \to \dots $$ Note that $\tilde H_n(Y) = H_n(Y)$ for $n > 0$. See see Hatcher's "Algebraic Topology" p.118 and Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$..
Now look at $$H_1(CA) \to H_1(X \cup CA) \to H_1(X \cup CA, CA) \to \tilde H_0(CA) \to \tilde H_0(X \cup CA) \to H_0(X \cup CA,CA) \to 0 $$ Since $H_1(CA) = \tilde H_0(CA) = 0$, you get the desired result.