Contradiction in Computation of Homology Groups of the Mapping Class Group of a Surface?

Question

One of the two main results of a paper by Nathalie Wahl on homological stability of the mapping class group of a surface is the following:

Theorem 1.2 The map $H_*(\delta_g) : H_*(\Gamma_{g,1};\mathbb{Z}) \to H_*(\Gamma_{g,0};\mathbb{Z})$ is surjective for $* \leq \frac23g+1$ and an isomorphism for $* \leq \frac23g$.

Here, $\Gamma_{g,r} := \pi_0\text{Diff}(S_{g,r} \, \text{rel} \, \partial)$ denotes the mapping class group of a surface $S_{g,r}$ with genus $g$ and $r$ boundary components such that all diffeomorphisms fix the boundary of the surface. Moreover, the map $\delta_g : \Gamma_{g,1} \to \Gamma_{g,0}$ is induced by gluing a disk to the one boundary component in $S_{g,1}$.

There is a much earlier paper by Harer, the main result of which (in our setting) is the following:

Theorem A Consider $\Gamma_{g,r}$ as above. Let $n$ be the number of distinguished points in $S_{g,r}$, i.e. points which are fixed by diffeomorphisms of the surface. Then we have $$ H_2(\Gamma_{g,r}) \cong \begin{cases} \mathbb{Z}^{n+1} & g \geq 5, \; r+n > 0\\ \mathbb{Z} \oplus \mathbb{Z}/(2g-2)\mathbb{Z} & g \geq 5, \;r=n=0 \end{cases} $$

In the setting of Wahl's paper, I believe we have $n=0$, since the author never mentions any distinguished points. Now, if we fix some $g \geq 5$ then we know that the map $H_2(\delta_g)$ is an isomorphism $H_2(\Gamma_{g,1}) \to H_2(\Gamma_{g,0})$, since the inequality $2 \leq \frac23 g$ is true by the assumption on $g$. However, using Theorem A, this translates to an isomorphism $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}/(2g-2)\mathbb{Z}$, since $n=0$, which is a contradiction, as the torsion is non-trivial.

Question: What exactly goes wrong in this line of reasoning? Am I overlooking any details?

One idea I have is that perhaps Wahl assumes that we always have one distinguished point, i.e. $n=1$, or that we can assume without loss of generality that all diffeomorphisms of $S_{g,r}$ fix a single point (outside of the boundary $\partial S_{g,r}$ of course). This would at least be sufficient to resolve the contradiction.

Answer 1

Kindly Nathalie Wahl has pointed out that the contradiction can easily be resolved as follows: It turns out that Harer's result simply contained a mistake in the torsion part, and his paper was later corrected. A correct statement of the result is given in a paper by Korkmaz & Stipsicz, namely

Theorem 1.1 If $g \geq 4$ and $n,r \geq 0$, then $H_2(\Gamma_{g,r}^n) \cong \mathbb{Z}^{n+1}$.

In particular this covers our above case with $g \geq 5$, $r=1$ and $n=0$. Then Theorem 1.2. of Wahl's paper says we have an isomorphism $\mathbb{Z} \to \mathbb{Z}$, so no contradiction.