Computing a projection-induced map on homology of the surface of genus $g$ Let $M_g$ be obtained from a sphere $S$ by gluing $g$ tori to it along disks, in such a way that (if one allows for slight abuse of notation):
$M_g/S\cong \bigvee^gM_1$
Observe that $H_2(M_g)\cong \mathbb{Z}$, and $H_2(\bigvee^gM_1)\cong \mathbb{Z}^{g}$.
Let $\pi:M_g\rightarrow\bigvee^gM_1$ be the projection.
Then we have an induced map on homology $\pi_*:\mathbb{Z}\rightarrow \mathbb{Z^g}$.
My geometric intuition tells me this must be the diagonal map $n\mapsto (n,\ldots,n)$, but I am struggling to come up with a rigorous argument for this.
In particular, I've tried to construct generators represented by singular chains which interact nicely with $\pi$, but so far my attempts end up depending on geometric intuition anyway.
 A: It's not really precise to say "gluing $g$ tori to a sphere $S^2$". Instead, it's possible to state something like "attaching $g$ tori to $S=S^2\setminus\bigsqcup_{i=1}^gB_i^2$" (removing $g$ open disks from the the 2-sphere and attach tori to the boundary by a homeomorphism that identifies the boundary of the attaching tori to $S$ counterclockwise)
With this new definition of $S$, $(M_g,S)$ becomes a good pair, meaning that $H_p(M_g, S)\cong \tilde{H}_p(M_g/S)$. Also, note that $M_g/S\approx\bigvee_{i=1}^g T^2$.
The homology exact sequence says
\begin{align}0\to H_2(S)\overset{i_*}{\to} H_2(M_g)\overset{\pi_*}{\to}H_2(M_g,S)\overset{\partial_*}{\to} H_1(S)\overset{j_*}{\to} H_1(M_g)\to H_1(M_g,S)\\
\to H_0(S)\to H_0(M_g)\twoheadrightarrow H_0(M_g,S)\to 0
\end{align}
By definition, $S\simeq\bigvee_{i=1}^{g-1}S^1$, which implies $H_2(S)\cong 0$ and $H_1(S)\cong \bigoplus_{i=1}^{g-1}\Bbb Z$. Hence, $\pi_*$ is injective. Now, we need to look at the boundary map $\partial_*:\Bbb Z^{\oplus g}\to\Bbb Z^{\oplus g-1}$. This map is surjective because $j_*$ is trivial. If $\beta\in H_2(M_g,S)$ is a class of relative cycle, then $\partial_*(\beta)$ is the class $\partial\beta\in H_1(S)$. For the usual basis of $H_2(M_g, S)$, each $\beta_i\mapsto \alpha_i, i=2,...,g$, where $\alpha_i$ is a generator of $H_1(S)$, but $\beta_1\mapsto \alpha_1\simeq\sum_{i=2}^{g}(-\alpha_i)$ (the cycle enclosing the first removed disk is homologous to the combination of other generating cycles in $S$). So, $\partial_*(\sum_{i}\beta_i)=0\implies\ker(\partial_*)=\langle(\beta_1,\ldots, \beta_g)\rangle=\operatorname{im}(\pi_*)$, meaning that $\pi_*$ is the diagonal map.
