# Cellular homology of the products with torus and Klein bottle with the circle

In what follows $T$ is the torus and $K$ is the Klein bottle.

I am just looking at Hatcher's example calculation using cellular homology of $T \times S^1$ and $K \times S^1$. Hatcher's homology notes are available here (see page 142 for question, and the commutative diagram and corresponding notes on page 141 for the definition of $\Delta_{\alpha \beta}$)

Hatcher states (in regards to $T \times S^1$):

Each $\Delta_{\alpha \beta}$ maps the interiors of two opposite faces of the cube homeomorphically onto the complement of a point in the target $S^2$ and sends the remaining four faces to this point. Computing local degrees at the center points of the two opposite faces, we see that the local degree is 1 at one of these points and −1 at the other, since the restrictions of $\Delta_{\alpha \beta}$ to these two faces differ by a reﬂection of the boundary of the cube across the plane midway between them, and a reﬂection has degree −1.

At first glance, I would have said that the faces have degree two. Are they a reflection because the faces are oriented in the opposite direction?

-
You should pick another place to host the image, it seems. What are $T$ and $K$? I have a guess, but by being explicit you maximize the changes that you will not get random answers! Likewise, what is $\Delta_{\alpha\beta}$. Another way to deal with this would be to give precise references (page and verse) into Hatcher, and a link to the online book, thereby making it trivial for the people who could answer your question to know what you mean. – Mariano Suárez-Alvarez Apr 1 '11 at 4:19
@Mariano: thanks I have done. The image works fine for me on several browsers (and operating systems!) so I'm not sure why it is not working for you! – Juan S Apr 1 '11 at 4:26
ImageShack seems to have deleted your image, and replaced it with an ad banner. If you can, please reupload the image (or something equivalent) using the image upload button in the editor toolbar (which will upload it to Stack Exchange's imgur account). – Ilmari Karonen Aug 17 '15 at 16:49