# Fundamental Group of Wedge of Two Torus

This problem is from Kosniowski's book "A First Course in Algebraic Topology". I am going to follow the hint but it is not easy for me to find $$C_A$$ and $$C_B$$. Is it correct that $$C_A$$ and $$C_B$$ should be the torus minus interior of a disk? If so, I could not see how that space contractible to a point. Thank you.

• Note that every manifold is locally Euclidean, i.e., each point of an $n$-manifold has an open neighborhood homeomorphic to $\Bbb R^n$. So, $C_A$ will be a small open subset of $A$ containing the point $x_0$ and homeomorphic to $\Bbb R^2$. Commented Oct 15, 2021 at 8:17
• @SumantaDas Ah great, thank you for your comment. Do you know some works that don't need manifold? I haven't learned manifold yet in my courses Commented Oct 15, 2021 at 9:06
• Have you seen the construction of the torus as a quotient of $[0, 1] \times [0, 1]$ by identifying the vertical and horizontal boundaries? This should allow you to conclude the same fact. Commented Oct 15, 2021 at 9:38
• @Rushy yes I have, thank you, I will try to convince my self Commented Oct 15, 2021 at 11:20

If you take the torus as $$\mathbb{T}:=S^1\times S^{1}$$ and pick points $$y_{1},y_{2}\in S^{1}$$ such that $$x_{0}\notin \{y_{1}\}\times S^{1}$$ and $$x_{0}\notin S^{1}\times \{y_{2}\}$$ then $$X:=(S^{1}-\{y_{1}\})\times (S^{1}-\{y_{2}\})$$ is an open neighborhood of $$x_{0}$$ by definition of the product topology and is contractible because $$X$$ is homeomorphic to $$\mathbb{R}^{2}$$.