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Exercise $19.1.C.$ in Ravi Vakil's notes reads as follows, with the bold part indicating the part I'm having trouble with:

Suppose $\pi: X \rightarrow Y$ is a finite morphism whose degree at every point of Y is $0$ or $1$. Show that $\pi$ is injective on points (easy). If $p \in X$ is any point, show that $\pi$ induces an isomorphism of residue fields $\kappa(\pi(p))\rightarrow \kappa(p)$. Show that $\pi$ induces an injection of tangent spaces.

Where here the degree of $\pi$ at $y$ means that dimension of the global sections of $\pi^{-1}(y)$ as a $\kappa(y)$ vector space.

I can show that $\pi$ is injective on points, since the fibre over a point is a finite scheme over $\operatorname{Spec}$ of the residue field there and thus a finite, discrete set of points. Thus by the degree hypothesis, it is either empty or a single point with the same residue field, which is enough to prove injecttivity on points and that the residue fields of $p$ and $\pi(p)$ are isomorphic. I can't work out how to prove the injectivity on tangent spaces though.

I know it's equivalent to surjectivity on the level of cotangent spaces, which is the same as asking for the image of $\mathfrak{m}_{\pi(p)}$ in $\mathfrak{m}_{p}/\mathfrak{m}_p^2$ to be the whole thing. I've tried playing around with Nakayama's lemma and using that $\mathfrak{m}_{\pi(p)}$ and $\mathfrak{m}_{p}/\mathfrak{m}_p^2$ are both $\mathcal{O}_{Y,\pi(p)}$ modules to try to show that $\pi^{\sharp}(\mathfrak{m}_{\pi(p)})+\mathfrak{m}_p^2 = \mathfrak{m}_p$, but haven't had any luck so far and have run out of ideas.

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    Degree 1 at $\pi(p)$ implies $\pi^{-1}(\pi(p))=p$ and $k(p)=\mathcal{O}_{X,p}/\mathfrak{m}_{\pi(p)}\mathcal{O}_{X,p}$.2017-01-07
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    @Mohan Sure, I mentioned both of those things in the post. Could you please clarify how that gives injectivity on the tangent space though? Perhaps I'm missing something obvious?2017-01-07
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    Doesn't it say $\mathfrak{m}_{\pi(p)}\mathcal{O}_{X,p}=\mathfrak{m}_p$? This is what you wanted to show, isn't it?2017-01-07
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    @Mohan Oh, I guess I mis-read the subscript on the ideal in your comment, you're right, that's exactly what is needed, thanks!2017-01-07

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