Splitness of a short exact sequence on a curve Let $C$ be a curve with genus $g > 1$. Consider the product $C \times C$, with natural projections $p_1$ and $p_2$ (from the first and second factor, respectively) to $C$. 
Consider the following short exact sequence of sheaves on $C \times C$:
$$0 \rightarrow p_2^* K_C^{\otimes 2} \rightarrow p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta) \rightarrow K_C \rightarrow 0$$
where $K_C$ is the canonical bundle of $C$, $\Delta$ is the diagonal in $C \times C$, and the map $p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta) \rightarrow K_C$ is the restriction map to $\Delta$. The last entry $K_C$ is a line bundle on $\Delta$, which is a sheaf on $C \times C$ supported on $\Delta$.
Now consider the projection ${p_1}_*$ of this sequence. Is the resulting sequence on $C$ split?
 A: The answer is still negative, though I feel the previous answer is incorrect. Let $L=p_1^*K_C, M=p_2^*K_C$ and  then saying $0\to p_{1*}M^2\to p_{1*}M^2(\Delta)\to K_C\to 0$ (the last is surjective, since the corresponding $R^1$ is zero) splits is same as sayining after twisting by $K_C^{-1}$, the section $1$ can be lifted. This implies in particular $H^0(L^{-1}\otimes M^2(\Delta))\neq 0$. But $p_{2*}(L^{-1}\otimes \mathcal{O}(\Delta))=0$ since $g>1$ and thus the $H^0=0$ which proves that the sequence that you need is not split.
A: I came up with an alternative answer. Let $\mathcal{M}_{g,n}$ be the moduli space of Riemann surfaces of genus $g$ with $n$ marked points. Then our sequence
$$0 \rightarrow {p_1}_* p_2^* K_C^{\otimes 2} \rightarrow {p_1}_* (p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta)) \rightarrow K_C \rightarrow 0$$
is in fact the same as
$$0 \rightarrow H^0(C,K_C^2) \otimes \mathcal{O}_C \rightarrow T^*\mathcal{M}_{g,1}|_{C} \rightarrow K_C \rightarrow 0$$
which is dual to
$$0 \rightarrow T_C \rightarrow T\mathcal{M}_{g,1}|_{C} \rightarrow H^1(C,T_C) \otimes \mathcal{O}_C \rightarrow 0$$
This last sequence is maximally non-split.
