Let $(L,V)$ be a base point free $g^r_d$ on a curve $C$. Then we have an exact sequence
$0\rightarrow M_{L,V}\rightarrow\mathcal{O}_C^{r+1}\xrightarrow{\theta} L\rightarrow 0$,
where we just let $M_{L,V}$ to be the kernel of $\theta$. People usually consider this sequence as the pullback of the (first twist of) Euler sequence
$0\rightarrow \Omega_{\mathbb{P}^r}(1)\rightarrow \mathcal{O}_{\mathbb{P}^r}^{r+1}\rightarrow\mathcal{O}_{\mathbb{P}^r}(1)\rightarrow 0$
via the morphism $\varphi\colon C\rightarrow \mathbb{P}^r$, which is induced by $(L,V)$. I could not see that so far.
Pulling back the Euler sequence with $\varphi$ clearly yields the exact sequence,
$\varphi^*(\Omega_{\mathbb{P}^r}(1))\rightarrow\mathcal{O}_C^{r+1}\xrightarrow{\theta} L\rightarrow 0$. So showing exactness of the left most map would yield the claim, but I don't see why this should be the case. Any ideas?
Actually this is my general problem. If I pullback an exact sequence with a non-flat map then I have no clue about the exactness of the resulting sequence. What is a good way to analyze this problem in general? Considering the long exact sequence with Tor sheaves or trying to right down the explicit map and making a brute force calculation?
This is because the right hand term is locally free. Pullback is locally tensor product. So you can extend your exact sequence by a Tor term, namely $Tor_{O({\mathbb{P}^r})}$$(O_C,O_{\mathbb{P}^r}(1))$ which is 0 since $O_{\mathbb{P}^r}(1)$ is locally free.