# Restriction to any submanifold of its domain is still immersion.

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 $df_q(x) = 0$.

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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 2 '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 2 '13 at 15:48