Topological mapping class groups of 4-manifolds It is a classical result of Quinn that for a simply-connected 
closed $4$-manifold $X$ the isometries of its intersection form 
are in one-to-one correspondence with 
$\pi_0 \text{Homeo}(X)$. (Isotopy of 4-manifolds, 1986)
Let $X$ have some simple fundamental group, say $\mathbb{Z}_2$, 
and let $h\colon X \to X$ be a diffeomorphism which acts trivially 
on $H_2(X;\mathbb{Z})$.
Is $h$ isotopic to the identity? (through $\text{Homeo}(X)$)
Edit: let $\tilde{X}$ denote the universal cover of $X$, and let $s \colon \tilde{X} \to \tilde{X}$ be the covering involution so that $\tilde{X}/s \cong X$. One can take a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$ of $h$ (there are two, but take any of them.) Another reasonable assumption on $h$ is that we want $\tilde{h}$ to be isotopic to either $\text{id}$ or to $s$.
Edit: The second question has been removed.
 A: Let $X = (S^2\times S^2)/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $(x, y) \mapsto (-x, -y)$. Note that $H_2(X; \mathbb{Z}) \cong \mathbb{Z}_2$, so every diffeomorphism acts trivially. 
Consider the diffeomorphism $f : S^2\times S^2 \to S^2\times S^2$ given by $(x, y) \mapsto (x, -y)$. This descends to a diffeomorphism $g : (S^2\times S^2)/\mathbb{Z}_2 \to (S^2\times S^2)/\mathbb{Z}_2$. Letting $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ be the universal covering map, we have a commutative diagram
$$\require{AMScd}
\begin{CD}
S^2\times S^2 @>{f}>> S^2\times S^2\\
@V{\pi}VV @VV{\pi}V \\
(S^2\times S^2)/\mathbb{Z}_2 @>{g}>> (S^2\times S^2)/\mathbb{Z}_2
\end{CD}$$
Taking $\pi_2$ of this diagram, we get a commutative diagram of abelian groups
$$\require{AMScd}
\begin{CD}
\mathbb{Z}\oplus\mathbb{Z} @>{f_*}>> \mathbb{Z}\oplus\mathbb{Z}\\
@V{\pi_*}VV @VV{\pi_*}V \\
\mathbb{Z}\oplus\mathbb{Z} @>{g_*}>> \mathbb{Z}\oplus\mathbb{Z}
\end{CD}$$
Note that $\pi_* = \operatorname{id}$ as $\pi$ is a covering map, but $f_*$ is given by  $(a, b) \mapsto (a, -b)$. By commutativity, the same is true of $g_*$. In particular, $g_* \neq \operatorname{id}$ and therefore $g$ is not homotopic to the identity map. 
Alternatively, note that $(S^2\times S^2)/\mathbb{Z}$ is orientable. As $\pi\circ f = g\circ\pi$, the maps $f$ and $g$ have the same degree, and it is easy to see that $f$ has degree $-1$. Again we see that $g$ is not homotopic to the identity map.
Geometrically, $X = \operatorname{Gr}(2, 4)$, the Grassmannian of unoriented two-planes in $\mathbb{R}^4$, and $S^2\times S^2 = \operatorname{Gr}^+(2, 4)$ the corresponding oriented Grassmannian. The diffeomorphisms $f$, $g$ are the maps given by $P \mapsto P^{\perp}$.
