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In the construction of Floer homology, one shows a formula that connects the Maslov index/ Conley–Zehnder index $\mu$ with the dimension of the moduli spaces of connecting gradient flow lines: $$\dim \mathcal{M}(x,y) = \mu(x)-\mu(y).$$ In the final step for this formula, a closed symplectic path consisting of four separate curves is chosen, so that the index along each of them is constant — and then, one uses that the path is contractible, in order to show that the sum of these indices must equal zero, so one can solve for the one that is wanted. See e.g. Salamon - Lectures on Floer homology, "proof of theorem 2.2", p. 23. According to Salamon, it is obvious that this path is contractible in $SP(n)^*$, i.e. it has Maslov index $0$. Other sources also just state this fact, e.g. Weber - Topological Methods in the Quest for Periodic Orbits, p. 87.

But frankly, I don't really see why it should have this property — so I would be really glad if someone could give me a short answer or idea.

I know that this is a very specific question, but I would still be very happy if anyone could pitch in their ideas.

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    $\begingroup$ I didn't look at any further context but my reading for what Salamon writes is that there is a map [0,1] x [-T, T] to the relevant space of operators, and what you're interested is the boundary of this rectangle. The boundary of a rectangle is null-homotopic in said rectangle, so pushing it forward to the relevant space of operators it remains null-homotopic. Have I misunderstood what he wrote? $\endgroup$
    – mme
    Commented Jul 29, 2022 at 22:17
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    $\begingroup$ @mme Thanks for your answer! Is there really such a theorem? I mean, it is obvious that $\partial([0,1]\times [-T,T])$ is null-homotopic in $[0,1]\times [-T,T]$, but why should the image-path be null-homotopic as well? Even in the very basic case where the map is just an identity and maps $\partial([0,1]\times[-T,T])\subset [0,1]\times [-T,T]$ to $\partial([0,1]\times[-T,T])\subset ([0,1]\times [-T,T])\setminus\{0\}$, the image path is not contractible... or am I seeing it all from the wrong perspective? $\endgroup$
    – Martin
    Commented Jul 29, 2022 at 23:06
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    $\begingroup$ Yes, if $H$ is a null-homotopy $S^1 \times I \to X$ and $f: X \to Y$ is a continuous map, then the composite $fH: S^1 \times I \to X$ is a null-homotopy of the composite. In your second case there is no map from the rectangle to the punctured rectangle which is the identity on the boundary. $\endgroup$
    – mme
    Commented Jul 30, 2022 at 0:15

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