Computing relative homology group of $S^1 \vee S^1.$ 
Let $p$ be the wedge point of $S^1 \vee S^1.$ Compute $H_* (S^1 \vee S^1, S^1 \vee S^1 - p).$

Is there any convenient space $X$ such that $S^1 \vee S^1 - p$ deformation retracts to $X\ $? Then by long exact sequence of relative homology group and by using five lemma we can conclude that $$H_*(S^1 \vee S^1, S^1 \vee S^1 - p) \cong H_*(S^1 \vee S^1,X).$$
How do I find such $X.$ It is clear that $S^1 \vee S^1 - p$ is two disjoint copies of $1$ simplices each of which are contractible. Hence I think $S^1 \vee S^1 - p$ deformation retracts to two points. Am I right? But then how do I compute the homology group?
Any idea will be appreciated. Thanks for your time.
 A: To remove this question from the unsolved/unanswered queue, I am writing all my comments in this answer.
Let $X:=\Bbb S^1\lor \Bbb S^1$ and $A:=X\backslash p$ where $p$ is the wedge point. Consider the reduced homology long exact sequence of the pair $(X,A)$: $$\cdots\to \widetilde H_n(A)\to \widetilde H_n(X)\to H_n(X,A)\to \widetilde H_{n-1}(A)\to \cdots$$ Note that $A\cong (0,1)\sqcup (0,1)$, so $$\widetilde H_n(A)=\begin{cases}\Bbb Z&\text{ if }n=0,\\0&\text{ otherwise}.\end{cases}\text{ and } \widetilde H_n(X)=\begin{cases}\Bbb Z\oplus \Bbb Z&\text{ if }n=1,\\ 0&\text{ otherwise.}\end{cases}$$ Hence, $\color{red}{H_n(X,A)=0\text{ for }n\geq 2}$.
Now, $X$ is path-connected and $A$ is a non-empty subspace of $X$, so by this we have $\color{red} {H_0(X,A)=0}$.
Finally, we have to compute $H_1(X, A)$. Let $U$ be a small open contractible neighborhood of $p$ such that $U\backslash p\cong (0,1)\sqcup(0,1)\sqcup(0,1)\sqcup(0,1)$. Now, excising  the closed set $X\backslash U$ from the pair $(X,X\backslash p)$ we have the inclusion induced isomorphism $H_n(U,U\backslash p)\to H_n(X,X\backslash p)$ for each $n\geq 0$.
In particular, $H_1(U,U\backslash p)\simeq H_1(X,X\backslash p)$. Now, consider a part of the reduced homology long exact sequence of the pair $(U,U\backslash p)$: $$\cdots\to \widetilde H_1(U\backslash p)\to \widetilde H_1(U)\to H_1(U,U\backslash p)\to \widetilde H_0(U\backslash p)\to \widetilde H_0(U)\to H_0(U,U\backslash p)\to 0$$Now, $U$ is path connected implies $\widetilde H_0(U)=0$ and $U\backslash p$ has four path components implies $\widetilde H_0(U\backslash p)\simeq \Bbb Z\oplus \Bbb Z\oplus\Bbb Z$. Also, $U$ is contarctioble implies $\widetilde H_1(U)=0$. Therefore, $$\color{red}{H_1(X,A)=H_1(X,X\backslash p)\simeq H_1(U,U\backslash p)\simeq \widetilde H_0(U\backslash p)\simeq \Bbb Z\oplus \Bbb Z\oplus\Bbb Z.}$$
