Let $S_1, S_2$ be two topological manifolds without boundary and let $D_1,D_2$ be two disks embedded in $S_1$ and $S_2$ respectively. Define $X_i:=S_i\setminus\text{int}(D_i)$. Let $Y$ be the space by gluing $\partial D_1$ to $\partial D_2$. I want to show that $Y$ is topological manifold. I have trouble showing this for points on the boundary components. Let $x\in \partial D_1$. Then $x\sim \tilde{x}\in\partial D_2$. I want to find an open neihbourhood of $x$ homeomorphic to $\mathbb{R}^2$. How can I do this?
Gluing along boundary of manifold
general-topologymanifolds
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It is highly nontrivial that the connected sum of manifolds is well-defined. First, you actually want to assume that you can thicken your embeddings of discs (the term is "locally flat"); otherwise one of your discs might be one side of the Alexander horned sphere! You want to prove that every embedding of the disc $D^n \hookrightarrow M$ is isotopic (meaning that the two embeddings are homotopic through locally flat embeddings). Two reasonable exercises are to show that if $f$ and $g$ are embeddings, then $g$ is isotopic to an embedding whose image is contained in the interior of the image of $f$; and to prove that if all (locally flat) embeddings of discs are isotopic, then the connected sum is well-defined. You'll then want the "isotopy extension theorem" for locally flat embeddings.
Now what you want to invoke is the annulus theorem to actually construct the final isotopy. This is not easy! You'll note that the Wikipedia article implies that the connected sum of manifolds was only finally proved well-defined in 1982. (It's a bit easier to prove that the connected sum of smooth manifolds is well-defined.)
If you are cynical about such theorems and demand to see proofs before you use these results, the way you should interpret theorems about $M \# N$ is "For every connected sum of $M$ and $N$, ..."
Well, manifolds with corners are not one and the same as topological manifolds. Let me come back to that point later, and address the easy parts of your question first.
In fact $\partial C=(S^{1}\times [0,1])\cup (D^{2}\times\{0,1\})$ is indeed homeomorphic to $S^2$. One way to see this is to subdivide $S^2$ at two latitude lines: the arctic circle at $67^\circ$ north latitude and antarctic circle at $67^\circ$ south latitude. The required homeomorphism takes the northern polar cap above $67^\circ$ north latitude to $D^2 \times \{1\}$; it takes the southern polar cap to $D^2 \times \{0\}$; and it takes everything between the arctic and antarctic circles to $S^1 \times [0,1]$. You can write down a completely concrete formula for this homeomorphism using a piecewise smooth formula with three pieces, expressed in terms of spherical coordinates on $S^2$ and polar coordinates on $D^2$.
However, this does not give a manifold-with-corners structure to $S^2$: the "corner structure" might appear to be the arctic and antarctic circles; but those are circles, they are disjoint from the topological boundary (which is actually empty), and the only "corners" on a 2-manifold-with-corners are isolated points on the topological boundary.
In fact the only manifold-with-corners structure on $S^2$ is one whose charts have values in $(0,\infty)^2$, which is equivalent to an actual topological manifold structure (with empty boundary).
As for your prism structure, you simply miscounted the edges, there are 9 edges: three around the arctic circle; three cutting across from the arctic circle to the antarctic circle; and three around the antarctic circle.
So what about your one and the same comment? It helps here to think in terms of category theory. Intuitively, a manifold with corners has more structure than a topological manifold. By forgetting this structure, you get a forgetful functor from the category of manifolds with corners to the category of topological manifolds.
What exactly has been forgotten? You write that the model space $[0,\infty)^d$ for a manifold with corners is homeomorphic to the model space $[0,\infty) \times \mathbb R^{d-1}$. This is true. However, the definition of manifold with corners has a rather strong restriction on overlap maps, which you have not mentioned. If the $U,V \subset [0,\infty)^d$ are the open subsets that are the targets of two charts in your manifold-with-corners atlas, and if $\psi : U \to V$ is the overlap map itself, then not only must $\psi$ be a homeomorphism (from $U$ to $V$), but $\psi$ must respect the corner structures. I suggest that you read exactly what atlases are, focussing on the overlap condition, in whatever textbook you might have on this topic. But in brief: the map $\psi$ must take the origin in $U$ to the origin in $V$ (assuming $U$ or $V$ contains the origin); it must take the union of the coordinate axes (intersected with $U$) to the union of the coordinate axes (intersected with $V$); it must take the union of the coordinate 2-planes to the union of the coordinate 2-planes; and so on. All of this extra structure is forgotten when you pass to the category of topological manifolds.
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I'm going to assume that you know all about quotient spaces (if you don't then you'll have to learn that because otherwise no answer to your question will make sense).
Another thing you need is the collar neighborhood theorem from differential topology, applied to the boundaries of $X_1$ and $X_2$. That theorem says that there exist neighborhoods $N_1$ of $\partial X_1$ and $N_2$ of $\partial X_2$, and diffeomorphisms $f_i : N_i \to \partial X_i \times [0,1)$ (for each $i=1,2$) such that $f(x) = (x,0)$ for each $x \in \partial X_i$ .
Now choose a diffeomorphism $g : \partial X_1 \to \partial X_2$. And then we have the quotient topological space $Y$ together with the quotient map $q : X_1 \cup X_2 \to Y$ obtained by identifying each $x \in \partial X_1$ with $g(x) \in \partial X_2$, so $q(x)=q(g(x))$.
Let me alter your notation slightly: I'll choose $x_1 = \partial D_1 = \partial X_1$ and $x_2 = g(x_1) \in \partial D_2 = \partial X_2$, which corresponds to the point $x = [x_1] = [x_2] \in Y$ (here I use the notation $[\cdot]$ to denote the corresponding point in the quotient space; I'm unsure whether this is what you intend in your question when you put the $\tilde{}$ symbol over something).
So now I have to describe a manifold chart in $Y$ for the point $x$. To do this, I'll choose a manifold chart in $\partial X_1$ around $x_1$, i.e. an open subset $U_1 \subset \partial X_1$ containing $x_1$ and a diffeomorphism $\phi_1 : U_1 \to B$, where $B$ is the unit open ball in Euclidean space. From this I get a manifold chart in $\partial X_2$ around $x_2$, namely $U_2 = g(U_1)$ and $\phi_2 = \phi_1 \circ g^{-1} : X_2 \to B$.
In $Y$ define the open chart around $x$ as follows: $$U = q\bigl(f_1^{-1}(U_1 \times [0,1)) \cup f_2^{-1}(U_2 \times [0,1))\bigr) $$ The map $\psi : U \to B \times (-1,+1)$ will be given by the following formula. Each $y \in U$ has one of two forms, and we give the formula for each form:
This is well-defined if $y$ has both forms (which only happens when $t=0$).
And then, by tracing through all the definitions and using all the theorems you can possibly find from quotient spaces, it follows that $\psi$ is a homeomorphism from $U$ to $B \times (-1,+1)$.