# Is the inclusion map always smooth?

This maybe a very silly question. But I am quite confused.

If $M$ is a smooth manifold, and $A\subset M$ is a submanifold, then is the inclusion map $i:A\longrightarrow M$ smooth?

My guess is that it might not be smooth. Because the manifold structures might not be compatible. But I am not able to precisely say this. I will be happy if someone helps!

• What definition of submanifold do you use? In the ones I'm used to, if $A$ is a smooth submanifold of $M$, the inclusion is smooth. – Daniel Fischer Mar 10 '14 at 18:18
• So an immersion, not necessarily an embedding. But you have the smoothness explicitly given in the definition. – Daniel Fischer Mar 10 '14 at 18:51
• @John, that person is me! :) – gradstudent Mar 10 '14 at 18:53
• @Daniel: it says that the map is 1-1, so it's actually an embedding; immersion would be "locally 1-1", no? – John Hughes Mar 10 '14 at 18:54
• No but what if I have for $A\subset M$ (where $A$ has some manifold structure) an $f:A\longrightarrow M$, $C^\infty$ where $f$ is not the inclusion. $f$ could be some other weird one-one immersion right? I am asking in that case will $i$ still turn out to be $C^\infty$? – gradstudent Mar 10 '14 at 18:56

As explained in the comments, smoothness is guaranteed by the definition of submanifold. The concern that "the manifold structures might not be compatible" is reasonable, but submanifolds are defined so that compatibility holds. Here is a non-example: let $M$ be the curve $y=|x|$ in the plane, considered as the image of $\mathbb R$ under $f(x) = (x,|x|)$. We can make $M$ a smooth manifold by declaring $f$ a diffeomorphism. Now $f:\mathbb R\to M$ is smooth but $f:\mathbb R\to\mathbb R^2$ is not; the smooth structures are not compatible. But of course, $M$ is not a submanifold of $\mathbb R^2$ here.

The definition you have (which is the definition of an immersed submanifold) has the smoothness of the inclusion map built in.

• Oh thank you! Now it's getting clearer. A subset becomes a sub manifold when ee declare the inclusion to be a diffeomorphism, is what I understand now. – gradstudent Mar 11 '14 at 5:20