Return to Question

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.(Here $A$ is subset of $[n]$)

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and Pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

they write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially the second one (I think the last term is a sheaf not a linear space!)

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.(Here $A$ is subset of $[n]$)

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and Pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

They write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

In the paper they imply that second map is obtained by natural fiber evaluation $H^0(f^*{T_X}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially what is fiber evaluation map?

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and Pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

they write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially the second one (I think the last term is a sheaf not a linear space!)

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.(Here $A$ is subset of $[n]$)

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and Pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

they write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially the second one (I think the last term is a sheaf not a linear space!)

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

they write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially the second one (I think the last term is a sheaf not a linear space!)

Suppose that $K=D(A;B;D_1;D_2)$ be a boundry divisor of $\overline{M}_{0,n}(X,\beta)$ and let $\overline{M}_A=$ $\overline{M}_{0,A \cup \bullet}(X,\beta)$ where $\bullet$ is additional marking.let $e_A$ be evaluation map on $\bullet$.

Now let $\overline{M}_A(X,\bar{t_A})$ be $\bar{t_A}$-rigid moduli space.In FP-notes by Fulton and Pandharipande at page 31 they define differential of $e_A$ at $[f]$ obtain as below:

they write we have a natural linear maps: $Def(f)\to H^0(f^*{T_X/T_C(-\bullet)}) \to T_X(f(\bullet))$

I want to know why we have these two linear maps especially the second one (I think the last term is a sheaf not a linear space!)