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If $\phi:M\longrightarrow N$ is an injective smooth map between two manifolds, then is $d\phi_m:M_m\longrightarrow N_{\phi(m)}$, the induced map between the tangent spaces injective too?

I tried the following : If $v\in M_m$ is such that $d\phi_m(v)=0$, then for all $g$, $C^{\infty}$ function in a neighbourhood of $\phi(m)$, $d\phi_m(v)(g)=0$, that is $v(g\circ\phi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.

I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $\phi$ being injective will automatically make it an immersion.

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  • $\begingroup$ If it helps future readers, Tu says the prototype of an immersion is the inclusion map, which is both an immersion and injective. So I guess this particular injective map being the prototype of immersions is what may lead people to think that injective maps are immersions or that immersions are injective. $\endgroup$ – Selene Auckland Jun 27 at 9:49
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The answer is no. Let $\phi:\mathbb R\rightarrow\mathbb R^2$ be $\phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.

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    $\begingroup$ You can also use the function $f(x)=x^3$ from the real line to itself. $\endgroup$ – Moishe Kohan Feb 11 '14 at 10:33
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If the map $\phi:M\longrightarrow N$ is an diffeomorphism then $d\phi_m:M_m\longrightarrow N_{\phi(m)}$ is an isomorphism.

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