Disclaimer - I don't have much experience in topology/complex geometry, so I apologize if what I'm asking is too elementary for this site.

Let $S$ be some orientable surface obtained by removing finitely many points from a closed surface.

The Riemann moduli space for $S$ is just the set of isomorphism classes of complex structures. In other words, it's the quotient of the set of all complex structures on $S$, by the equivalence relation where two complex structures $\mu,\mu'$ on $S$ are equivalent if there is an homeomorphism $f : S\rightarrow S$ such that $\mu\circ f = \mu'$.

According to Wikipedia, the Teichmuller space for $S$ can be described as also the quotient of the set of all complex structures on $S$ by the same equivalence relation, except where the $f$ is required to be isotopic to the identity.

In Lochak's paper here: http://www.math.jussieu.fr/~lochak/textes/curves.pdf

he says that points of the Teichmuller space can be described by pairs $(X,f)$ where $X$ is a Riemann surface (ie, a complex structure on $S$), and $f : S\rightarrow X$ is a diffeomorphism, and that $(X,f)\sim(X',f')$ if $f'\circ f^{-1}$ is homotopic to an isomorphism (ie, a biholomorphic map) from $X\rightarrow X'$.

My first question is - In this case, is $f'\circ f^{-1}$ being homotopic to an isomorphism equivalent to it being isotopic to one?

My second question is - Why are these two definitions equivalent? I can't seem to see why having an isomorphism $X\rightarrow X'$ that is isotopic to the identity is equivalent to $f'\circ f^{-1}$ being homotopic to an isomorphism.

In fact this just seems false. Ie, consider the case where $X = X'$ as Riemann surfaces, and pick markings $f,f'$ such that $f'\circ f^{-1}$ is not isotopic to the identity, then the equivalence of the two definitions would seem to imply that there are isomorphisms between $X$ and $X'$ that are isotopic to the identity, and ones that are not, which seems strange to me.