Whether or not such a simple CW complex can be made a $C^{\infty}$ manifold? 
Problem  Let $X$ be the space obtained by attaching two disks to $S^1$, the first disc being attached by the 7 times around,i.e. $z \to z^7$, and the second by the 5 times around.
  Can $X$ be made a $C^{\infty}$ manifold? Why or why not?

I can calculate the homology group of $X$; and I try to find some information about it. I try to check some properties for manifolds, for instance, Poincaré duality. However, there seems no clear test to determine whether it can be made a manifold.
 A: Note that $H^2(X, \mathbb{Z})\cong\mathbb{Z}$. So if $X$ is a smooth manifold then it is an oriented compact surface. $X$ is not homeomorphic to $S^2$ because $X-\{\text{pt}\}$ is not contractible, while $S^2-\{\text{pt}\}$ is. So $X$ should have nontrivial first cohomology by the classification of oriented compact surfaces. But $H^1(X, \mathbb{Z})=0$. So $X$ cannot be a smooth manifold
A: Part of the theory of spheres and manifolds contains the the fact that for any $k$-manifold $M$ and any point $p \in M$, the relative homology gorup $H_k(M,M-p;\mathbb{Z})$ is isomorphic to $\mathbb{Z}$.
However, if you take your complex $X$ and take $p$ to lie on the circle $S^1$ to which the two discs have been attached, a short computation using excision shows that $H_k(X,X-p;\mathbb{Z})$ is zero in all dimensions except $k=2$ and $H_2(X,X-p;\mathbb{Z}) \approx \mathbb{Z}^{11}$. That's an "eleven", as in $11 = 7 + 5 - 1$.
Added later: Here are some details on the computation of the relative $H_2$ term. 
By construction, the point $p$ has a closed neighborhood $Y$ which is obtained from a closed segment $I$ by attaching 7 closed half discs to $I$ and then attaching $5$ more closed half discs to $I$, for a total of 12 half discs; each of these attachments identifies $I$ homeomorphically with the straight side of the half-disc. Applying excision we have
$$H_2(X,X-p;\mathbb{Z}) \approx H_2(Y,Y-p;\mathbb{Z})
$$
The subset $Y-p$ deformation retracts to its frontier in $X$, which I'll denote $\partial Y$, and so
$$ H_2(Y,Y-p;\mathbb{Z}) \approx H_2(Y,\partial Y;\mathbb{Z})
$$
Since $Y$ itself is contractible, using the long exact sequence of the pair $(Y,\partial Y)$ we obtain
$$H_2(Y,\partial Y;\mathbb{Z}) \approx H_1(\partial Y;\mathbb{Z}) \approx \mathbb{Z}^{\beta_1}
$$
where $\beta_1$ is the first Betti number of $\partial Y$.
The space $\partial Y$ is a connected graph with two vertices and $7+5 = 12$ edges, so the $0^{\text{th}}$ Betti number equals $1$ and applying the Euler characteristic we have
$$\beta_0 - \beta_1 = 2 - 12
$$
$$\beta_1 = 1 - 2 + 12 = 12 - 1 = 11
$$
A similar computation shows that $H_k(X,X-p;\mathbb{Z})$ is zero if $k \ne 2$.
