I was wondering if the set of singular loops is a (somewhere) submanifold of loop space? The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space).
I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space?
EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.) 
So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)? 
EDIT 2. I reformulate question. $map(S^1 \to \mathbb R^3)$ is a functional space, so we can apply a technique of singularity theory. Generic map $f$ with one self-intersection has one-parameter versal deformation $V:[0,1]\times S^1\to \mathbb R^3$ - and any deformation are induced from $V$. Does it imply that $D$(set of singular loops) near the $f$ is a submanifold of codimension 1 in sense of Fréchet or Frolicher? 
ADDED. Andrew Stacey explains in his answer and in http://ncatlab.org/nlab/show/on+the+manifold+structure+of+singular+loops why a stratum of a loop space is not a submanifold (the reason is tha same as in a standart smooth injection of line to plane where image is not a submanifold).
But locally, as Ryan said, each stratum is a submanifold (in sense of Frechet). And the kast question is:
For a smooth submanifold $X⊂Y$ of codimension 2, for a general point $x∈X$ we always have a map from small neghbourhood $U$ of $x$ to $D^2$ ($x∈U⊂Y,f:U→D^2$) such that $U\cap X= f^{−1}(0)$. I belive that in situation of space of loops there is no such map... Is it true?
 A: My interpretation of your question is that you would like to see a stratification of the space of smooth maps $S^1\to M^n$.  It suffices to consider maps $\newcommand{\r}{\mathbb R} \r \to \r^n$, and for simplicity I will take $n=3$.
I think this is a very interesting question.  I suspect the answer is no, there is not a nice stratification "all the way down", but that you have to go to fairly high codimension before you run into problems.  (See the answers to this question and the references therein.)
In particular, I don't think Ryan's example presents a problem.  We just have to recognize that the set of maps with a single double point (and no other singularities) consists of several strata: a codimension 1 stratum where the double point is transverse, a codimension 3 stratum where the tangents at the double point are colinear but the higher derivatives are in general position, and higher codimension strata where the two curves are tangent to higher order.
More specifically, a generic colinear double point is (after change of coordinates in the domain and range) of the form $[t \mapsto (t, 0, 0); t \mapsto (t, t^2, t^3)]$.  I claim that singularities of this form are a codimension 3 submanifold of the space of all smooth maps.  As in Ryan's answer, a 3-dimensional space of perturbations transverse to this stratum has the form $t\mapsto (t, t^2+a, t^3+bt + c)$.
The link of this stratum is a 2-sphere with an embedded "figure 8".  The nonsingular parts of the figure 8 correspond to single transverse double points.  The central (singular) point of the figure 8 corresponds to a pair of transverse double points.  Thus there is no reasonable way to make the subset of maps with a single not-necessarily-transverse double point into a nice stratified set, much less into a manifold.  I think this was the point of Ryan's answer.  The point of my answer is that the space of all smooth maps nevertheless has a perfectly nice stratification near this sort of singularity.
The list of indecomposable strata for maps $\r\to \r^3$ starts out like this:


*

*codimension 1: single transverse double point

*codim 2: zero derivative with 2nd, 3rd and 4th derivatives linearly independent, i.e. conjugate to $t\mapsto (t^2, t^3, t^4)$

*codim 3: (a) triple point with tangents linearly independent; (b) double point with tangents colinear and higher derivatives independent (as above)

*codim 4: (a) triple point with tangents coplanar; (b) double point with one of the tangents zero (and higher derivatives generic); (c) colinear double point with some higher derivatives non-generic (like $[t\mapsto (t, 0, 0); t\mapsto (t, t^2, t^4)]$; not sure about this one)


Items from the above list can of course be combined.  For example, two generic double points and four generic triple points has codimension $2\cdot 1+ 4\cdot3 = 14$.
It would be nice if one could continue the above list so that the complement of all the strata had infinite codimension.  I'm not sure whether that's possible.
A: Let $L(M)$ be the space of smooth maps from $S^1$ to $M$. 
Let $D(M)$ be the subspace of $L(M)$ consisting of immersions $f : S^1 \to M$ where $f$ admits a unique pair of points $p,q \in S^1$, $p \neq q$ with $f(p)=f(q)$. 
My understanding of your question, is you want to know if $D(M) \subset L(M)$ is a submanifold.
I think it's not. 
"Most" (*) of $D(M)$ has co-dimension $m-2$ in $L(M)$, where $dim(M)=m$.  But the problem is $D(M)$ is a stratified space.  $D(M)$ has a subspace where $f'(p)$ and $f'(q)$ are linearly dependent.  Call this $D'(M)$.  The reason for the asterix is that $D(M) \setminus D'(M)$ has co-dimension $m-2$ in $L(M)$. This is a genuine submanifold, and you get "charts" by doing the Vassiliev resolution-of-singularities, as $D(M) \setminus D'(M)$ is just immersions that have a single regular double point. 
So the problem is getting charts around points of $D'(M)$.  They don't exist.  You can make the argument in general, but let's just take $M = \mathbb R^3$ just to make it somewhat concrete.   Take a local picture that looks like the graph of the functions:
$$ t \longmapsto (t,0,0)$$
i.e. the x-axis
and 
$$ t \longmapsto (t,t^2,0) $$
So this these are meant to be conjugate to two intervals of a map $f : S^1 \to \mathbb R^3$, describing a quadratic-type double point. 
We consider a little neighbourhood of $f$ by doing a small perturbation to $f$ in $L(M)$.
$$ t \longmapsto (t,t^2+a, bx+c)$$
This is the original map when $a=b=c=0$.  Its in $D'(M)$ only for that original map.  For $a>0$ there are no singularities at all, but for $a<0$ there are the singularities corresponding to all parameters that satisfy
$$\pm b\sqrt{-a}+c=0$$
but the central singularity, where $b=c=0$ is one with two regular double points. 
So this means that if $f \in D'(M)$, there are no manifold charts for $D(M)$ in $L(M)$ centered around $f$, since there are four "sheets" of $D(M) \setminus D'(M)$ incident to these points.  $D(M)$ isn't locally flat about these points. 
edit: Here is how you argue that $D(M) \setminus D'(M)$ is a submanifold of $L(M)$.  Given a point $f \in D(M) \setminus D'(M)$ it is an immersion $S^1 \to M$ which is an embedding except for a single regular double point, meaning $f(p)=f(q)$ for $p,q \in S^1$, $p \neq q$, and $f'(p)$ and $f'(q)$ are linearly independent.  The tangent space to $f$ is the sections of the pull-back $f^o(TM)$ bundle.  The tangent space to $D(M) \setminus D'(M)$ at $f$ is given by certain constrained tangent vector fields along $f$.  Thinking of $f^o(TM)$ as the tangent space to $L(M)$, it's the subspace of $f^o(TM)$ such that if $p,q \in S^1$ are the double points, then the vector over $p$ and the vector over $q$ are both tangent vectors in $T_{f(p)}M$, and we demand that the components of these vectors normal to $img(Df_p) \oplus img(Df_q)$ agree.  By design, this is a $(m-2)$-dimensional constraint.  The Vassiliev resolution of singularities idea -- pushing off the double-point -- is the complementary space to the tangent space of $D(M)\setminus D'(M)$ in $L(M)$.  So this is a splitting of $T_f(L(M))$ into $T_f(D(M)\setminus D'(M)$ direct sum with an $(m-1)$-dimensional space corresponding to the deformations that remove the singularity. This also gives you your submanifold charts for $D(M)\setminus D'(M)$ at $f$. 
Note: I meant to use standard notation for pull-backs but raising an asterix in math mode does strange things.  So I use $f^o(bundle)$ instead. 
Technically, what I've described in "edit" is a splitting of the tangent space to $L(M)$ at a point $f \in D(M) \setminus D'(M)$.  The sub-manifold chart for $D(M) \setminus D'(M)$ is precisely the same chart as you use for $L(M)$ at $f$, but pre-composed with this splitting isomorphism.  That this map is a submanifold chart for $D(M)\setminus D'(M)$ amounts to the observation that you can break this flow up into an isotopy that carries $img(Df_p) \oplus img(Df_q)$, together with a motion of the two arcs in this 2-dimensional space (which isn't an isotopy, it's just a 1-parameter family of immersions). 
A: I started answering this question, as much to understand how to fit Torsten and Ryan's arguments into what I already know, but the answer got a bit long.  Since I find the input format here a little constricting, I ended up writing it as an nLab page and have posted it there.  Here's the summary:


*

*Torsten and Ryan are correct: none of the variants of the spaces that you have described (to date) can be thought of as a submanifold of the smooth loop space.  In my answer I try to expand on these objections a little, but the core of the objections are those that they describe.

*In particular, Ryan's objection is based on the fact that the space of loops with a exactly one coincidence point is not an open subset of the space of loops with at least one coincidence point.  (Contrast this with the fact that the space of loops with no coincidence points is a submanifold of the space of all loops.)

*There is a smooth manifold nearby.  It is the space of singular loops with a marked coincidence point.  (One could also take, as Ryan says, the subspace of singular loops with a single coincidence point and linearly indepdendent tangent vectors at that point.)

*In considering how to go from a marked loop to an unmarked one, we can see (hopefully clearly) why we do not get a manifold.
The nlab page is at: http://ncatlab.org/nlab/show/on+the+manifold+structure+of+singular+loops (though I think that the title would be better as "on the manifold structure of singular loops, or lack thereof").
A: The following argument I think should work to prove that the answer is no (though I haven't checked whether it really fits in with the proper definition): Consider a loop where three points of $S^1$ maps to the same point in $M$ but with linearly independent (to be on the safe side) tangents. Then you keep within the singular loops by moving two of the three points in the same way but the third in any which way. To first order that should correspond to a $TM$-valued vector field on $S^1$ which is the same at the two points but arbitrary at the third. However, there are three possibilities for the choice of the two points that should travel together and all the vector fields that you get from varying the choices will not form a linear subspace. Hence what you get is rather three submanifolds (corresponding to the three choices) coming together meeting transversally.
