I have been trying to prove that the torus $T$ with any point$\{x\}$ removed is homotopy equivalent to the wedge sum of two circles. Now I have the intuitive idea of the result, I can see how we can continuously deform $T\setminus \{x\}$ into $S^1\vee S^1$. I have found that this question has been asked before, but the answers use diagrams to prove the statement. But if I wanted a concrete proof of the fact how would I go about it?
From the definition of homotopy equivalence we want two continuous maps $f:T\setminus \{x\}\to S^1\vee S^1$ and $g:S^1\vee S^1\to T\setminus \{x\}$ such that $f\circ g$ and $g\circ f$ are both homotopic to the identity. I believe this $f,g$ should be somewhat canonical, but I can't guess what I should write here. Can someone point me in the right direction?
This is a difficult homotopy equivalence to visualize, but I've attempted to draw a picture of what's going on. First you poke a hole in your surface starting at
Best Answer
Write the torus as a closed square with opposite edges identified. Take the square's centre as the point to be removed. Deform the square minus its centre radially to its perimeter. When we identify the edges this becomes a deformation retraction of the torus minus a point to a bouquet of two circles.