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I'm trying to work with the space $X=S^1\times I$.

It is obvious that $X\simeq S^1$ and therefore $H_1(X)=H_1(S^1)=\mathbb Z$, but I want the properties of $X$ itself.

I would assume that a generator of $X$ would be one loop around $X$, like $S^1 \times {0}$. But I can't understand why it would be like that -

every $f:\:I\to X$ is either null-homotopic (and in that case it is obvious that $f\in image(\partial _2 )$ that is it is a boundary, or it is similar to the generator we have chosen - a positive number of loops around $X$. it is possible for the image of this loop to not intersect the image of the generator.

my question is how would we show that $f$ in this is case is a multiplication of the generator? that is probably similar to why two loops, say $h=S^1\times{0}$ and $g=S^1\times {1}$ have $h-f$ is a boundary of a 2-triangle. we could connect their starting points with a line, but then why would the surface be a continuous image of a 2-triangle?

hope i made myself clear.

Stefan Hamcke
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Dargich
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  • A generator of $H_1(S^1 \times [0, 1])$ is indeed a loop without a basepoint in $S^1 \times [0, 1]$. In fact, for any path-connected space $X$, $\pi_1/[\pi_1, \pi_1] \cong H_1$, obtained from the homomorphism sending a loop in $\pi_1(X, x_0)$ to $H_1(X)$ by forgetting about the basepoint. – Balarka Sen May 16 '15 at 14:11
  • Also, this question is closely related to the very first question I ever asked on this site: http://math.stackexchange.com/questions/50362/homology-why-is-a-cycle-a-boundary – Ben Blum-Smith May 16 '15 at 15:00

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Let's concentrate on your case: you are trying to find an explicit homology between $h$ and $g$.

Draw the line $0\times I$ which connects the two loops' basepoints. If you cut along this line the space becomes a square; draw one diagonal of this square. (Don't actually make the cut! I'm not trying to alter your space. I just invoked it to describe the diagonal I wanted you to draw!)

We have now partitioned the surface into 2 triangles; these two triangles are images of maps of standard $2$-simplices into your space $X$, the sum of which is an element of the 2-chain complex. If you choose the orientations correctly, its boundary will be exactly $g-h$; the rest of the boundary will cancel. (You have to choose orientations so that cancellation happens along the line and diagonal I just asked you to draw.)

Ben Blum-Smith
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  • Thank you Ben. This is actually what I thought about myself, but I can't quite understand the part about cutting. We do treat $S^1\times {0,1} \cup 0\times I$ as if it was cut along $0\times I$, so it does moderate the space. I mean, how could it be a continuous image of the 2-simplex otherwise? – Dargich May 16 '15 at 14:25
  • Now that I think about it, shrinking things into a point or a line is still continuous, isn't it? just not homemorphic. Am I right? – Dargich May 16 '15 at 14:34
  • Yes, exactly. E.g. the map from the square to the cylinder obtained by gluing a pair of opposite edges together is a continuous map, but it doesn't have an inverse. – Ben Blum-Smith May 16 '15 at 14:43
  • This kind of continuous map that is not invertible is really important to understand homology. For example, how do we know that a loop $k$ that goes twice around $S^1\times 0$ is homologous to $2h$? Because $2h-k$ is the boundary of a map from a single triangle into $X$ that squishes the whole triangle into $S^1\times 0$. – Ben Blum-Smith May 16 '15 at 14:47