Let $C$ be an affine, integral curve and $f: \widetilde{C} \to C$ be its normalization. Let $g:D \to C$ be a finite, affine, surjective morphism (note $D$ need not be reduced, but can assume generically reduced). Denote by $D'$ the base change of $\widetilde{C}$ by the morphism $g$ and $g': D' \to D$ the resulting morphism. Is the morphism $g'$ schemetheoretically surjective i.e., the induced ring homomorphism $\mathcal{O}_D \to \mathcal{O}_{D'}$ is injective?
No. Let $A \to B$ be a ring map. Let $M$ be a finite $A$module such that $M \to M \otimes_A B$ is not injective. Then with $A' = A \oplus M$ where $M$ is an ideal of square zero and $B' = A' \otimes_A B$ the base change, we see that $A' \to B'$ is not injective.
Apply this with $A = k[t^2, t^3]$ where $k$ is a field, $B = k[t]$ is the normalization of $A$ and $M = k[t^2, t^3]/I$ where $I$ is the ideal generated over $A$ by $t^5$. Then $t^6 \not \in I$ but $M \otimes_A B = B/J$ where $J$ is the ideal in $B$ generated by $t^5$ and so $t^6 \in J$. Hence $M \to M \otimes_A B$ is not injective.

$\begingroup$ Sorry, forgot to add the condition that the morphism $g$ is surjective. I have edited the question. $\endgroup$ – Jana May 27 at 13:53

$\begingroup$ The morphism $\text{Spec}(A \oplus M) \to \text{Spec}(A)$ is surjective as it has a section. This is my last comment on this question. $\endgroup$ – Johan May 27 at 19:46

$\begingroup$ I am sorry, I do not think your example works. Ofcourse $M \to M \otimes_A B$ is not injective. But, this does not imply $A' \to B'$ is not injective. In particular, in your example you claim that the morphism $k[t] \to (k[t^2, t^3] \oplus k[t^2, t^3]/(t^5)) \otimes k[t]$ is not injective, where the tensor product is over $k[t^2, t^3]$. But, note that, $t^a$ for any $a$ maps to $(1 \oplus 1) \otimes t^a=(t^a \oplus t^a) \otimes 1$ which is never zero. I am assuming that the ring homomorphism from $k[t^2, t^3] \to k[t^2, t^3] \otimes k[t^2, t^3]/(t^5)$ sends $f$ to $(f,f \mod t^5)$. $\endgroup$ – Jana May 27 at 21:11

$\begingroup$ @Jana you consider the map $A \to B'$ instead of the map $A' \to B'$. $\endgroup$ – Joshua Mundinger Jun 27 at 1:17

$\begingroup$ @JoshuaMundinger Sorry, I wrote it in a confusing manner, but you can use the arguments to check that the map from $A'$ to $B'$ is injective. The important point to note is that as $g$ is assumed to be surjective, the most natural map from $A$ to $A \oplus M$ is the one that takes $f$ to $(f,f \mod t^5)$ (I think this is the map that Johan considers). But, for such a map, the induced map from $A'$ to $B'$ is injective. Probably, you can come up with a different injective map from $A$ to $A \oplus M$, under which $A' \to B'$ is not injective, but I could not think of such a map. $\endgroup$ – Jana Jun 27 at 6:25