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I was able to show they are both homotopy equivalent to $S^1 \wedge S^1$. I cannot find any reason why they would not be homeomorphic, but I also cannot see any way to construct an explicit map. Any hints would be appreciated.

Emilio Minichiello
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    A standard way to prove such statements is to use cohomology with compact support: $H^1_c(T^2- p)\cong Z^2$, $H^1_c(S^2-{p_1,p_2,p_3})=0$. Cohomology with compact support is a topological invariant but not a homotopy-invariant. (It is only invariant under proper homotopy-equivalence.) – Moishe Kohan Jun 10 '19 at 17:21

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When you take a loop $\gamma$ which goes once around the torus $\Bbb T^2$ (and doesn't go on the point $p$ which was removed) then the complement of $\gamma$ in $\Bbb T^2-p$ is still connected.

The complement of any loop $\gamma$ in $\Bbb S^2-\{p_1,p_2,p_3\}$ is disconnected (because of Jordan curve theorem basically). Therefore they can't be homeomorphic.

Adam Chalumeau
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