The title is the question. Sorry, this isn't quite research level. I imagine the answer is wellknown, just not to me. Thanks for any help!

$\begingroup$ some tag with "topology" would be natural (I'm not sure whether this is considered as algebraic or geometric topology, or both) $\endgroup$– YCorCommented May 7, 2015 at 22:28

1$\begingroup$ btw there was no comment about the meaning of the question: does it mean continuous immersions modulo continuous isotopy, smooth immersion modulo smooth isotopy, and are these two point of views equivalent in any dimension? $\endgroup$– YCorCommented May 7, 2015 at 22:30

$\begingroup$ @YCor, I think "immersion" is inherently a smooth notion, so it means smooth immersions modulo smooth isotopy. But the SmaleHirsch theorem turns it into a topological problem about continuous maps. $\endgroup$– Dylan ThurstonCommented May 8, 2015 at 0:40

$\begingroup$ @DylanThurston: an immersion can be characterized as a map between manifolds $f:X\to Y$ such that for every $x\in X$ there are neighborhoods $V,V'$ of $x$ and $f(x)$ such that $f(V)\subset V'$ and the restriction of $V$ can be described in coordinates as the inclusion of a subspace of $\mathbf{R}^d$ into $\mathbf{R}^n$. This makes sense in the continuous setting. But OK, I understand there's consensus that the setting is the smooth setting. $\endgroup$– YCorCommented May 8, 2015 at 8:01

$\begingroup$ @YCor: this is smooth immersions up to 1parameter families of immersions. Sometimes this is called "regular isotopy" to distinguish it from "isotopy". We're talking about the homotopytype of the space of immersions. Immersions up to plain isotopy have an enormous amount of components, primarily indexed by the image (as a stratified space). $\endgroup$– Ryan BudneyCommented May 8, 2015 at 15:13
4 Answers
I'm not a professional topologist by any means, but let me give this a shot. There's some discussion in the first few lectures of John Francis from this course. Please correct me if I've made mistakes below...
By Smale and Hirsch, the space of immersions of $S^n$ into $R^{n+1}$ is homotopy equivalent to the space of unbased maps from $S^n$ into $V_{n}(\mathbb{R}^{n+1})$, $Map(S^n,V_n(\mathbb{R}^{n+1})$. Here $V_n(\mathbb{R}^{n+1})$ is the Stiefel manifold of $n$ frames in $n+1$space. $V_{n}(\mathbb{R}^{n+1})$ is homeomorphic to $SO(n+1)$.
See Ryan's answer for some more details.
Your question is then basically equivalent to characterizing $\pi_1(Map(S^{n},SO(n+1)))$, the fundamental group of that mapping space.
As B.S. points out in his answer, since $SO(n+1)$ is a group, $Map(S^n,SO(n+1))$ is homotopy equivalent to $SO(n+1)\times\Omega^n(SO(n+1))$, therefore the above fundamental group is $\pi_1(SO(n+1))\times\pi_1(\Omega^n(SO(n+1)))\cong\pi_1(SO(n+1))\times\pi_{n+1}(SO(n+1))$.
Though B.S. has pointed out that we can see that it's always nontrivial just from the first factor, we can in fact compute the group from known results on homotopy groups of $SO(n+1)$ (see e.g. this table compiled from the literature by Klaus Johannson).
The result for all $n$ is (scroll right in the grey box below):
n 1  2  3  4  5  6  8s1  8s  8s+1  8s+2  8s+3  8s+4  8s+5  8s+6 
                            
 Z  Z_2+Z  Z_2+Z_2+Z_2  Z_2+Z_2  Z_2  Z_2+Z  Z_2+Z_2+Z_2+Z_2  Z_2+Z_2+Z_2  Z_2+Z+Z_2  Z_2+Z_2  Z_2+Z_2+Z_2  Z_2+Z_2  Z_2+Z_4  Z_2+Z 
In the rightmost columns, $s$ is any integer greater than or equal to 1, and the plus signs denote direct sum. Apologies for the ugly formatting of the table.

1$\begingroup$ I think that is not what HirschSmale says here. HirschSmale says that the space of immersions is homotopy equivalent to the space of sections of the fiber bundle over $S^n$ with fiber $V_n(\mathbb{R}^{n+1})$ associated to the tangent bundle, which is trivializable only for $n = 1, 3, 7$. You also need to be more careful about the distinction between spaces of based vs. unbased maps; e.g. based maps from $S^1$ is the based loop space but free maps from $S^1$ is the free loop space, and these have different homotopy groups in general. $\endgroup$ Commented May 7, 2015 at 18:10

1$\begingroup$ @Qiaochu: literally you're right but there is a reduction in this particular case. SmaleHirsch says immersions (via the derivative) are homotopyequivalent to the bundle monomorphisms $TS^n \to T \mathbb R^{n+1}$, but using the almosttrivializability of $TS^n$ you can check this mapping space has the same homotopytype as $\Omega^n SO_{n+1}$. $\endgroup$ Commented May 7, 2015 at 18:14

4$\begingroup$ @Ryan: huh. Okay, I'm willing to believe that it has the same homotopy type as the space of maps from $S^n$ to $SO(n+1)$, but surely this doesn't in turn have the same homotopy type as the space of based maps...? $\endgroup$ Commented May 7, 2015 at 18:16

2$\begingroup$ Oh, yes, you're right there. It's the unbased loop space. The based loop space would be immersions with a fixed behaviour at a point. I'll supply the details in a short partialanswer. $\endgroup$ Commented May 7, 2015 at 18:17

2$\begingroup$ I think there's a fibration $\Omega^n(SO(n+1))\rightarrow X \rightarrow SO(n+1)$, where X is the space of unbased maps from $S^n$ to $SO(n+1)$. So you can look at the long exact sequence attached to this fibration to try to compute $\pi_1(X)$. Since $\pi_2(SO(n+1))$ vanishes, this only has five nontrivial terms. $\star \rightarrow \pi_{n+1}(SO(n+1)) \rightarrow \pi_1(X) \rightarrow \mathbb{Z}_2 \rightarrow \pi_n(SO(n+1)) \rightarrow \pi_0(X) \rightarrow \star$. By Kervaire's table for $\pi_{n+1}(SO(n+1))$ you basically have the answer up to knowing how $\pi_1$ acts on $\pi_n$ for $SO(n+1)$ $\endgroup$ Commented May 7, 2015 at 20:12
Your space is never simply connected (for $n\geq 1$). As already answered, it is (weakly) homotopy equivalent by SmaleHirsch hprinciple to the space of unbased maps from $S^n$ to $SO(n+1)$, which is itself $SO(n+1)\times \Omega^n SO(n+1)$ ($\Omega^n$ = based maps from $S^n$). So $\pi_1$ is at least $\pi_1(SO(n+1))$ (and more in general). In fact it is not even connected as soon as $\pi_n SO(n+1)$ isn't trivial, which occurs for $n=1,3,4,5$ and many more (maybe all except $n=2$?). But it is connected for $n=2$, which awarded its celebrity to Smale (sphere eversion), even if his advisor Raoul Bott didn't believed it at first. EDIT: according to j.c. $\pi_n SO(n+1)$ is trivial only for $n=2,6$. Quite an interesting fact !

1$\begingroup$ Your answer shows that for any choice of basepoint there is a canonical surjective homomorphism from this fundamental group to the cyclic group of order two (induced by the projection to $SO(n+1)$). This should correspond to a natural 2fold covering (of the space of immersions). What is it? $\endgroup$– YCorCommented May 7, 2015 at 22:26

$\begingroup$ Could you say a little more about why the space of unbased maps from $S^n$ to $SO(n+1)$ the same as $SO(n+1)\times\Omega^nSO(n+1)$? By the way, $\pi_n SO(n+1)$ is given in the table cited in my answer, it's nontrivial except for $n=2$ and $n=6$. $\endgroup$– j.c.Commented May 7, 2015 at 22:31

2$\begingroup$ @j.c.: Since $SO(n+1)$ is a group, the space of maps from $S^n$ to $SO(n+1)$ is the same as the product of $SO(n+1)$ (image of the base point of $S^n$) and $\Omega^n SO(n+1)$ (based maps $\infty\mapsto I$). By the way, I didn't know of the case $n=6$. Thanks ! $\endgroup$– BS.Commented May 7, 2015 at 23:26
Let me just fillin the gap in j.c.'s exposition. SmaleHirsch states that the derivative from the space of immersions $Imm(S^n, \mathbb R^{n+1})$ to the space of bundle monomorphisms $Mono(TS^n, T\mathbb R^{n+1})$ is a homotopyequivalence.
There is a cute observation that allows you to naildown the space of fibrewise onetoone maps $TS^n \to T\mathbb R^{n+1}$.
Notice that $TS^n$ is virtually trivial, i.e. $TS^n \oplus \epsilon^1 = S^n \times \mathbb R^{n+1}$, where $\epsilon^1$ is a line bundle over $S^n$.
So the idea is to extend any bundle monomorphism to an orientationpreserving bundle monomorphism $S^n \times \mathbb R^{n+1} \to T \mathbb R^{n+1}$. You can do this continuously, over the entire space using the a crossproduct type construction. This is a homotopyequivalence between $Mono(TS^n, T\mathbb R^{n+1})$ and the $n$fold free loop space of $SO_{n+1}$.

1$\begingroup$ The space of unbased maps from $S^n$ to a space is not the $n$fold free loop space; that would be the space of unbased maps from $T^n$. $\endgroup$ Commented May 7, 2015 at 18:45

2$\begingroup$ Qiaochu you're too much a grad student. That's needlessly pedantic. $\endgroup$ Commented May 7, 2015 at 19:38

6$\begingroup$ @QiaochuYuan If it helps, I suspect Ryan's comment lies in the space of unbased critiques from Professors to Grad Students. $\endgroup$ Commented May 7, 2015 at 21:11

1$\begingroup$ Unbased, only up to a controlled homotopy. $\endgroup$ Commented May 7, 2015 at 22:05

2$\begingroup$ @RyanBudney Observe: the mapping space $\text{map}(S^n,G)$ of a Lie group $G$ splits as a product $\Omega^n G \times G$. $\endgroup$ Commented May 8, 2015 at 2:02
For $n=1$, the space $Imm(S^1,\mathbb R^2)$ has $\mathbb Z$ many connected components described by the rotation index. In each case the fundamental group is $\mathbb Z$. See Thm 2.10 of here for the components with rotation index $\ne 0$, and see this paper for rotation index 0.