Alternative conditions for a function $\phi:S_g\rightarrow S_g$ to be homotopic to identity Let $S_g$ be a closed oriented surface of genus $g\ge 1$ and  $f:[0,1]\times S_g\rightarrow S_g$ be a function such that:


*

*$f(0,x)=x$ for every $x\in S_g$

*for every fixed $t\in [0,1]$ the function $f(t,\cdot):S_g\rightarrow S_g$ which to every $x\in S_g$ associates $f(t,x)\in S_g$ is continuous

*for every fixed $x\in S_g$ the function $f(\cdot,x):[0,1]\rightarrow S_g$ which to every $t\in [0,1]$ associates $f(t,x)$ is continuous (every point is moved along a continuous path)
Question: is $f(1,\cdot): S_g\rightarrow S_g$ homotopic to the identity?
It seems to me that the answer could be "yes", because the continuous paths "lock" the homotopy class of $f(1,\cdot)$, but I can not formalize it. However, in general a separately continuous function is not necessarily continuous and so the function $f:[0,1]\times S_g\rightarrow S_g$ could be not continuous and consequently it can not be considered as an homotopy between the identity and $f(1,\cdot)$.
If the answer to the previous question is "no", then is it at least true that there exists a $\overline t>0$ such that $f(\overline t,\cdot):S_g\rightarrow S_g$ is homotopic to identity?
 A: Neat question!  Let me start by giving a counterexample for $S^1$ instead of a surface.  Define $g:[0,1]\times[0,1]\to \mathbb{R}$ as follows.  First, we define $g(0,x)=x$ for all $x$.  Now suppose $t>0$.  For $0\leq x\leq t/2$, $g(t,x)=x$.  For $t/2\leq x\leq t$, $g(t,x)$ travels linearly from $t/2$ to $t-1$.  For $t\leq x\leq 1$, $g(t,x)=x-1$.
We now compose $g$ with the quotient map $p:\mathbb{R}\to \mathbb{R}/\mathbb{Z}= S^1$ and observe that $pg(t,0)=pg(t,1)$ for all $t$.  We can thus identify $x=0$ with $x=1$ in the domain of $pg$ and obtain a map $f:[0,1]\times S^1\to S^1$.
This map $f$ now satisfies your conditions.  It is clear that $f(0,x)=x$ and $f(t,\cdot)$ is continuous.  Continuity of $f(\cdot,x)$ is only potentially a problem at $t=0$, but there is no discontinuity since for any fixed $x$, $f(t,x)=x$ for all $t$ sufficiently close to $0$ (though how close is "sufficiently close" depends on $x$).
However, for any $t>0$, the map $f(t,\cdot)$ is nullhomotopic, and in particular is not homotopic to the identity.  Indeed, this is immediate from the fact that when we lift $f(t,\cdot)$ to a path in $\mathbb{R}$, we get $g(t,\cdot)$, and $g(t,0)=g(t,1)=0$.

Now, having built this example for $S^1$, you can adjust it to work for surfaces.  For instance, you could get an example for $S_0=S^2$ by just suspending the map $f(t,\cdot)$ for each $t$.  Or you could get an example for $S_1=S^1\times S^1$ by using the identity on one coordinate and the function above on the other.  For genus greater than $1$, I don't think you can directly construct an example from the example for $S^1$, but I think it should be possible to adapt the idea, though I haven't worked out the details.
