We were asked to compute the homology for the double torus, $X$, and a circle around one of the loops, $B$, of the torus (not a circle between the two halves of the torus) and were told that this space is homotopy equivalent to $T^2\lor S^1$. From here, I could compute the homology, but I was just wondering how you would show that $X/B$ is homotopy equivalent to $T^2\lor S^1$. The picture I got was a torus with two horns attached at a point.

Also, is there a more general construction here? So if I have $\#^gT^2$ can I kill of $k<n$ loops like $B$ to get a space $Y$ homotopy equivalent to $\#^{g-k}T^2\lor S^k$?

  • $\begingroup$ Are you sure about that? Isn't your space homotopy equivalent to $S^2 \vee S^1$? $\endgroup$ May 2, 2016 at 16:05
  • $\begingroup$ It can't be $S^2\vee S^1$, @NajibIdrissi. The LES of the pair $(X,B)$ reveals that $H_1(X,B)=\Bbb Z^3$. $\endgroup$ May 2, 2016 at 16:32

1 Answer 1


Your picture of two horns attached at a point is correct - but from there you can stretch the attached section until you have two small horns joined by a line, and from there deform it until it's $T \vee S^1$. I've attached a hastily-drawn picture.

In answer to your "general construction" question: hopefully that's more obvious now that you've seen how the construction works! You can kill $k < n$ loops to get a space homotopy equivalent to $\#^{g-k}T^2 \vee \bigvee _1^k S^1$

$X/B$ being homotoped to $T \vee S^1$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .