TITLE: Verification of a proof regarding the connected sum of two surfaces
QUESTION [0 upvotes]: I am trying to solve the following exercise:

Let $X_1, X_2$ be two surfaces. Lets consider charts $\varphi_j: U_j \to \mathbb{R}^2$ with $U_j \subset X_j$, $j= 1, 2$ and let $B_j = \varphi^{-1}_j(B_1(0)) \subset X_j, ~ X_j^0 = X_j - B_j$, $j= 1,2$. Prove that the connected sum $X_1 \# X_2$, defined as $(X_1^0 \sqcup X_2^0)/\sim$, where $x_1 \sim x_2$ if $x_j \in S_j = \varphi^{-1}_j(\partial B_1(0)) = \partial X_j^0, ~ \varphi_1(x_1)= \varphi_2(x_2) $, is homeomorphic to the space:
$$ (X_1^0 \sqcup (S_1 \times [ 0,1 ]) \sqcup X_2^0)/\approx $$
with $x_1 \in S_1$, $x_1 \approx (\varphi_1(x_1),0)$, and $x_2 \in S_2$, $x_2 \approx (\varphi_2 (x_2), 1)$.

The idea behind the exercise is very clear to me. It is the same to join the surfaces directly than doing it by means of a cylinder (because the cylinder with the top layer and the bottom layer ($C$) is homeomorphic to the sphere that is the unitary element of the connected sum).
But I don know if my proof is completely rigorous. Is this a valid proof of the above exercise? I would appreciate any tips and suggestions.

We want to prove that $ \frac{(X_1^0 \sqcup (S_1 \times [ 0,1 ]) \sqcup X_2^0)}{\approx} = X_1 \# C \# X_2$ and then the result follows because the connected sum is associative.
The previous statement holds because $$\frac{(X_1^0 \sqcup (S_1 \times [ 0,1 ]) \sqcup X_2^0)}{\approx} = \frac{\frac{X^{0}_{1} \sqcup C}{\sim} \sqcup X^0_2}{\sim}  $$
To finish the proof we can see that the cylinder (with the top and bottom lids) being homeomorphic to $S^2$ implies $X_1 \# C \cong X_1$

REPLY [0 votes]: I sent the above proof (with a more detailed explanation and a picture to illustrate my ideas) to my teacher and he accepted it as a rigorous proof, so I can answer this question saying that it is, in fact, a valid proof of that exercise.
{"set_name": "stack_exchange", "score": 0, "question_id": 1476690}