Given the surface of genus $g$, $M_g$, obtain $H_{\ast}(M_g)$ without using it's a $CW$-complex and knowing the homology of the Torus and $\pi_1(M_g)$.
I thought considering the space $A\subseteq M_g$ such that is one of the Torus $T$ inside $M_g$ and doing this exact sequence inductively :
$$\widetilde{H_2}(T-C)\longrightarrow \widetilde{H_2}(X)\longrightarrow H_2(X,T-C)\longrightarrow \widetilde{H_1}(T-C)\longrightarrow \widetilde{H_1}(X)\longrightarrow H_1(X,T-C)\longrightarrow \widetilde{H_0}(T-C)\longrightarrow 0$$
Since $\widetilde{H_0}\oplus \mathbb{Z} =H_0$ for any space, $\widetilde{H_0}(T-C)=0$. Also I have that: $$H_{\ast}(X,T-C)=\widetilde{H_{\ast}}(X/(T-C))\cong \widetilde{H_{\ast}}(M_{g-1})$$ And finally considering $T-C\cong S^1 \vee S^1$, so $\widetilde{H_2}(T-C)\cong 0$, $\widetilde{H_1}(T-C)=\mathbb{Z}\oplus \mathbb{Z}$ and $\widetilde{H_0}(T-C)=\mathbb{Z}$. Then the first long exact sequence becomes:
$$0\longrightarrow \widetilde{H_2}(X)\longrightarrow \widetilde{H_2}(M_{g-1})\longrightarrow \mathbb{Z}\oplus\mathbb{Z}\longrightarrow \bigoplus_{i=1}^{2g}\mathbb{Z}\longrightarrow H_1(M_{g-1})\longrightarrow 0$$ That is: $$0\longrightarrow \widetilde{H_2}(X)\longrightarrow \widetilde{H_2}(M_{g-1})\xrightarrow{i} \mathbb{Z}\oplus\mathbb{Z}\xrightarrow{g} \bigoplus_{i=1}^{2g}\mathbb{Z}\xrightarrow{f} \bigoplus_{i=1}^{2(g-1)}\mathbb{Z}\xrightarrow{0} 0$$
Since it's a long exact sequence, $\text{Im}(g)=\text{ker}(f)$, and by the first isomorphism theorem $\bigoplus_{i=1}^{2(g-1)}\mathbb{Z}=\frac{\bigoplus_{i=1}^{2g}\mathbb{Z}}{\text{ker}(f)}=\frac{\bigoplus_{i=1}^{2g}\mathbb{Z}}{\text{Im}(g)}$ so $\text{Im}(g)=\mathbb{Z}\oplus\mathbb{Z}$, obtaining that $g$ is an injective inclusion, so $\text{ker}(g)=0$, implying $\text{Im}(i)=0$, so $$\widetilde{H_2}(X)\cong \widetilde{H_2}(M_{g-1})$$ $$\Rightarrow H_2(X)\cong H_2(M_{g-1})$$ $$\Rightarrow H_2(X)\cong H_2(M_{g-1})\cong\dots \cong H_2(M_1)=H_2(T)=\mathbb{Z}$$ So
$$ H_{\ast}(M_g) = \begin{cases} \mathbb{Z} & \ast=0,2\\ \bigoplus_{i=1}^{2g}\mathbb{Z} & \ast=1\\ 0 & \ast>2 \end{cases} $$
Is this proof correct or I could have done it shorter? Essentially with knowing it's correct or not and pointing how to correct the mistakes I'm fine.