I'd like to show the following (intuitively clear) fact:
Given a torus $T^2$ and an embedded disk $D\subset T^2$ (put a disk in the middle of the square whose edges we identify to get the torus), consider $X=T^2\setminus int(D)$.
Claim: X does not retract to $\partial D\subset X$
What I tried: If there was such a retraction, we'd get in homology: $H_i(\partial D)\rightarrow H_i(X)\rightarrow H_i(\partial D)$. For i=1, this comes down to $\mathbb{Z}\rightarrow H_i(X)\rightarrow \mathbb{Z}$. The first map is induced by inclusion, the second by our retraction, so the composition is homotopic to the identity, so it induces an isomorphism. That means I have to compute the first homology group of X, which is where I'm stuck formally.
Is what I've done so far okay (or is there a better way?) and how do I finish the proof?
Best Answer
Hint 1: Removing the disc means you can strong deformation retract $X$ to the edges of the square, so you can calculate the homology of just the 1-skeleton.
Hint 2: You'll need to compute more than just the first homology group of $X$. To complete the proof you'll need to use information about the map of homology groups induced by the inclusion $\partial D \rightarrow X$.