1

If $f$ is an immersion, prove its restriction to any submanifold of its domain is an immersion.

Consider a submanifold $\tilde{X}$ of $X$, and take any point $p \in \tilde{X}$. Then when $d\tilde{f}_p(\tilde{x}) = 0$...

I was not able to show $\tilde{x} = 0$ here. Can I reduce this problem to the canonical submersion? In other words, if $d\tilde{f}_p(\tilde{x}) = 0$, then $df_q(x) = 0$ where $q = (p_1, \dots, p_n, 0 \dots, 0)$, and $x = (\tilde{x}_1, \dots, \tilde{x}_n, 0, \dots, 0$). And because $f$ is an immersion, $df_q(x) = 0$ implies $x=0$ and therefore $\tilde{x} = 0$.

Though, I am not quite comfortable with the interchange between manifold and submanifold. Specifically, I don't know if I can claim if $d\tilde{f}_p(\tilde{x}) = 0$, then $df_q(x) = 0$.

Michael Albanese
  • 99,526
1LiterTears
  • 4,572
  • 2
    Since $\tilde f$ is the restriction of $f$, what does this tell you about the relation between $d\tilde f$ and $df$? – Lee Mosher Jul 02 '13 at 15:44
  • Not sure - I guess $f$ carries $(x_1, ..., x_n, 0, ..., 0)$ to $(y_1, ..., y_m)$ while $\tilde{f}$ carries $(x_1, ..., x_n)$ to $(y_1, ..., y_n)$? So $df$ carries $(x_1, ..., x_n, 0, ..., 0)$ to $(y_1, ..., y_m)$ while $d\tilde{f}$ carries $(x_1, ..., x_n)$ to $(y_1, ..., y_n)$? – 1LiterTears Jul 02 '13 at 15:48

1 Answers1

5

Let $S$ be a submanifold of $M$, and $i : S \to M$ be the inclusion. If $s \in S$, then $di_s : T_sS \to T_sM$ is the inclusion map. As $f|_S = f\circ i$, we have

$$d(f|_S)_s = d(f\circ i)_s = df_{i(s)}\circ di_s = df_s\circ di_s.$$

As $di_s$ and $df_s$ are injective, $d(f|_S)_s$ is injective so $f|_S$ is an immersion.

Michael Albanese
  • 99,526