$T \times S^1$ and $K \times S^1$ (a question from Hatcher)

Note that 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 $K \times S^1$.

is the image to have in mind (the left is $T \times S^1$ the right is $K \times S^1$)

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?

Edit: 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 $\Delta_{\alpha \beta}$)

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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