I was reading *The symplectic Floer homology of a Dehn twist* by P. Seidel, which you can find [here](https://www.intlpress.com/site/pub/files/_fulltext/journals/mrl/1996/0003/0006/MRL-1996-0003-0006-a010.pdf).


In Lemma 3(ii) the following topological property of Dehn twists is stated without proof:



> 
> Let $\Sigma$ be an oriented compact surface. Let $C\_1,\dots,C\_n \subset \Sigma$ be noncontractible disjoint circles contained in $\Sigma$, and for every $i=1,\dots,n$ let $T\_i$ be a positively oriented Dehn twist around $C\_i$. Also, for every $i=1,\dots,n$ choose a sign $\sigma\_i \in \{\pm 1\}$. Assume that if a component of $\Sigma - \bigcup C\_i$ is a cylinder bounded by $C\_i$ and $C\_j$, then $\sigma\_i = \sigma\_j$. Let $T := T\_1^{\sigma\_1} \circ \dots \circ T\_n^{\sigma\_n}$.   
> 
> If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$, and $T \circ \gamma$ is homotopic to $\gamma$ relative to the endpoints, then $\gamma$ is homotopic relative to the endpoints to a path $\gamma'$ contained in $\mathrm{Fix} T$.
> 
> 
> 


I guess this is an (almost) trivial fact for experts on MCG of surfaces, hence the lack of proof, but could someone provide me a reference where this is proven?


**Remark.** For his purposes, Seidel assumes that $\Sigma$ has no boundary and that its genus is greater than 1, but I think the claim is still true in this generality.


Thank you in advance for your time.



Reference for a property of Dehn twists
The [Primer](https://press.princeton.edu/books/hardcover/9780691147949/a-primer-on-mapping-class-groups-pms-49) has "model proofs" for various steps of the proof.
[Bleiler's notes on Casson's lectures](https://www.cambridge.org/core/books/automorphisms-of-surfaces-after-nielsen-and-thurston/2AD58B246E36B971CCB92BD5B923BDB9) are also very good on these topics. (The notes cover far less material than the Primer, so they are much much shorter.)




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I find the given statement confusing, because the conclusion is not as strong as it should be. I think the conclusion should be as follows:



> 
> If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$,
> and if $T \circ \gamma$ is homotopic to $\gamma$ (rel endpoints), then there is a homotopy of $\gamma$ (rel endpoints) making $\gamma$ disjoint from the union $\cup C\_i$.
> 
> 
> 


**Proof:** We prove the contrapositive. Homotope $\gamma$ (rel endpoints) to have minimal intersection number with $\cup C\_i$. Suppose that the result still meets $\cup C\_i$. So $\gamma$ crosses one of the curves, say $C$. Let $(B\_j)\_j$ be the subcollection of $(C\_i)\_i$ which are all parallel to $C$. Let $X \subset \Sigma$ be an annulus that contains all of the $B\_j$ as essential curves in its interior. We may arrange matters so that $\gamma$ has minimal intersection number with $\partial X$. Let $\delta$ be a component of $\gamma \cap X$ which meets some of the $B\_j$. We may (and do) assume that the endpoints of $\delta$ are fixed by all of the $T\_i$.


Let $\Sigma^C$ be the cover of $\Sigma$ which is homeomorphic to an annulus and where $C$ does not unwrap. We use the initial point of $\delta$ as our base-point; let $\gamma^C$ be the resulting lift of $\gamma$ to $\Sigma^C$. Let $\delta^C \subset \gamma^C$ be the resulting lift of $\delta$. The arc $\delta^C$ crosses the lifts of (the correct) $B\_j$ due to our choice of base-point.


We now lift $T(\gamma)$ using the *same* base-point. It lifts (up to homotopies in $\Sigma^C$ supported in bigons disjoint from $C$) to a copy of $T\_C^n(\gamma^C)$. Here $n$ is the number of parallel copies of $C$ crossed by $\delta$ (and the sign depends on the sign of the twists along the $B\_j$). Since $T\_C^n(\gamma^C)$ has different winding number (than $\gamma^C$) with the curves $B\_j$, it follows that they are not homotopic (rel endpoints) in $\Sigma^C$. Thus (by homotopy lifting) $T(\gamma)$ and $\gamma$ are not homotopic (rel endpoints) in $\Sigma$, as desired.


