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I'm learning the concepts of immersions at the moment. However I'm a bit confused when they define an immersion as a function $f: X\rightarrow Y$ where $X$ and $Y$ are manifolds with dim$X <$ dim$Y$ such that $df_x: T_x(X)\rightarrow T_y(Y)$ is injective.

I was wondering why don't we let $f$ be injective and say that's the best case we can get for the condition dim$X <$ dim$Y$(since under this condition we can't apply the inverse function theorem)?

Also does injectivity of $df_x$ inply the injectivity of $f$ (it seems that I can't prove it)?

How should we picture immersion as (something like the tangent space of $X$ always "immerses" into the tangent space of $Y$)?

Thanks for everyone's help!

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"Does injectivity of $\mathrm{d}f_x$ imply the injectivity of $f$?" Consider the standard covering map $f:\mathbb{R}^1 \to \mathbb{S}^1$... – Willie Wong May 27 '13 at 11:01
"How should we picture...?" Locally immersions are imbeddings. Near a point $p$ in the domain and near $f(p)$ in the image, there are coordinates under which the (restriction of the) immersion is the standard imbedding of Euclidean space into a Euclidean space of equal or higher dimension: $(x_1,..., x_m) \mapsto (x_1,...,x_m, 0,...,0)$. There is an important subtelty here: the image of this restriction may not be open in the image of the whole domain. This is what distinguishes imbeddings from injective immersions. – Tim kinsella May 27 '13 at 11:53
up vote 2 down vote accepted

Think of a particle moving around a figure 8 with nowhere zero speed. This parametric curve gives you an immersion $f\colon\mathbb R \to\mathbb R^2$ that is not injective. If you restrict the domain to make it a bijection (which you can do), the image is not a submanifold but is called an immersed submanifold.

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Hi Prof. Shifrin! Thanks a lot for your great answer! – Evariste Jun 1 '13 at 5:51

I think a glimpse on the wikipedia's article helps. Immersions usually are not injective because the image can appear "knotted" in the target space.

I do not think you need $\dim X<\dim Y$ in general. You can define it for $\dim X=\dim Y$, it is only because we need $df_{p}$ to have rank equal to $\dim X$ that made you "need" $\dim X\le \dim Y$.

I think you can find in classical differential topology/manifold books (like Boothby's ) that an immersion is locally an injection map $$\mathbb{R}^{n}\times \{0\}\rightarrow \mathbb{R}^{m+n}$$You can attempt to prove this via inverse function theorem or implicit function theorem. The proof is quite standard.

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Thanks a lot! Great answer! – Evariste Jun 1 '13 at 5:51

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