Topology: crosscaps and handles I am struggling with the following exercise:

Let $A$ be a compact connected surface. $\quad$ We say that a crosscap $\otimes$ is inserted into $A$ if an open disc $D$ is removed from $A$ and opposite points of the resulting circular boundary $\partial D$ are identified.$ \quad $We say that a handle is attached to $A$ if two open discs are removed from $A$ and their boundaries are identified. $\quad$ Now, show that:


a) inserting a crosscap into $A$ is equivalent to gluing a Moebius strip into the hole resulting from removing a disc,


b) attaching a handle to $A$ is equivalent to forming the connected sum $A \# T^2$.

Unfortunately, I have no idea how to show this. Furthermore, I don't even know what is meant by "gluing a Moebius strip into the hole". Gluing it in in what way?
 A: This can all be done without recourse to the classification of surfaces or the Seifert-van Kampen theorem. Instead, I would say that a good understanding of this problem is an important step in (some treatments of) the proof of the classification of surfaces.
For (a), let's use $p : A - D \to F$ to be the quotient map. Also, let $D'$ be a larger open disc whose interior contains $D$ concentrically. Consider the boundary circles $C' = \partial D'$ and $C = \partial D$, and let $R$ be the compact annulus in $A$ with $\partial R = C' \cup C$.
Question: What is the image of $R$ under the quotient map $p$?
Answer: That image $p(R)$ is homeomorphic to the quotient space of $R$ where opposite points of $C$ are identified, and one can show by direct construction  that it is homeomorphic to a Möbius band which I'll denote $M = p(R)$. Thus, removing $D$ and gluing opposite boundary points of $C$ results in the same surface as removing $D'$ and gluing in the Möbius band $M = p(A)$ by identifying the circle $p(C') \subset p(A)$ with the boundary of $M$ by a homeomorphism.
For (b) the formulation is not quite correct: one needs to be careful with orientations. Let me assume that $A$ is orientable, in which case one needs to choose an orientation on $A$ which induces an orientation on $A-(D_1 \cup D_2)$ and thereby induces boundary orientations on each circle $C_i = \partial D_i$, and then one needs to require that the gluing homeomorphism $C_1 \mapsto C_2$ reverses boundary orientations. The danger is that if you instead preserve boundary orientations then the result is $A \# K$, where $K$ is the Klein bottle. (The case that $A$ is nonorientable is more complicated, leading to the conclusion that $A\#T$ and $A\#K$ are homeomorphic).
So, with that proviso about orientations, there is a similar answer but one needs a preliminary construction. Let $D_1,D_2$ be the two open discs that are removed, and denote their boundary circles $C_i = \partial D_i$. What you need next is that there exists an open disc $D \subset A$ which "engulfs" both of $D_1,D_2$, in the sense that in an appropriate coordinate system on $D$, the discs $D_1,D_2$ are disjoint round subdiscs of $D$. One way to see this, using path connectivity, is to construct an embedded path $\gamma$ in $A$ whose intersection with $D_i$ is a single point of $C_i$; one can then take $D$ to be a regular neighborhood of $D_1 \cup \gamma \cup D_2$, which one can show by direct construction is homeomorphic to a disc with subdiscs as said.
Once you have $D$, now let $P$ be the compact "pants" subsurface of $D$ with $\partial P = \partial D \cup \partial D_1 \cup \partial D_2$. One can show, again by direct construction, that when the quotient map $p : A - (D_1 \cup D_2) \to F$ is restricted to $P$, the image $p(P)$ is a torus.
