Let $X$ and $Y$ be smooth algebraic varieties over $mathbb{C}$. Let $f$ be a continuous map from complex points of $X$ to $Y$. Are there Zariski opens $U$ and $V$ inside $Xtimes mathbb{A}^1$ and $Ytimes mathbb{A}^1$ respectively such that $U$ contains $Xtimes{0}$ and $Xtimes {1}$ (similarly $V$ contains $Ytimes {0}$ and $Ytimes {1}$) with the property that a continuous map $g$ from $U$ to $V$ can be assigned to $f$ in a way that $g_{Xtimes {0}}= f$ and $g_{Xtimes {1}}$ is a regular morphism?
Tag: homotopic
algebraic topology – CW structure from map homotopic to the identity?
Suppose $X$ is a $CW$ complex, and $h:Xto X$ is a (cellular) map which is homotopic to the identity.
What can one say about the the image $h(X)$? In particular I am hoping to give it some natural CW structure, hopefully homotopic to X.
As an example: if $A$ is a contractible subcomplex, applying the homotopy extension property gives a map $h:Xto X$ with $h(A)= point$, and the image of $h$ is ‘morally’ $X/A$ and homotopic to $X$, although this is a bit tricky to prove, and involves $A$ being contractible.
If $X$ is $k$connected, the inclusion of the $k$skeleton is nullhomotopic, we can apply the homotopy extension principle to get $h:Xto X$, and I would really like to conclude $X/X^{k}$ is homotopic to $X$. This is not true as written, but $X$ is homotopic to a complex with trivial $k$skeleton. I know this can be proved with Whitehead’s theorem, but can someone help me get a more ”barebones” picture along the lines above?
algebraic topology – Map that is homotopic to the identity on $S^n$
I’m looking for a homotopy that satisfies the following:
Let $B subset S^n$ be an open ball around the north pole $Nin S^n$. There exists a homotopy $$ F : S^n times I to S^n$$
with $F(x,0) = x$ for all $x in S^n$ and $g(x) = F(x,1)$ satisfies
$$g(B) = S^n – {N}, g(S^n – {N}) = S, g(N) = N, text{g is smooth},$$ where $S$ is the south pole.
I have an idea of how $g$ should look like. I should kind of stretch the open ball $B$ onto all of $S^n – {N}$ if that makes sense. So points inside the ball $B$ should go towards the south pole on great circles between the north and south pole. But I have no idea how to write it down nicely and I’m also not sure how the homotopy should look like. I’d appreciate any help!
differential geometry – Homotopic paths in a simply connected space
In my algebraic topology course, we define a simply connected topological space as a topological space $X$ whose fondamental group $pi_1(X)$ is reduces to the trivial group ${1}$ and my teacher told me that in such a space, if we consider two paths $gamma_1:(0,1) rightarrow X$, $gamma_2:(0,1) rightarrow X$ such that $gamma_1(0) = gamma_2(0) = x_0$ and $gamma_1(1) = gamma_2(1) = x$, then $gamma_1, gamma_2$ are homotopic.
I do not really understand why this is true. Of course, my first intuition was to concatenate these paths such that
begin{equation*}
gamma(t) =
begin{cases}
gamma_1(2t) &text{for } t in (0,1/2)\
gamma_2(2(1 – t)) & text{for }t in (1/2,1)
end{cases}
end{equation*}
and by simply connectedness of $X$, we conclude that $gamma$ is homotopic to the constant loop, but from here I do not see how to conclude that $gamma_1$ and $gamma_2$ are homotopic. Could someone give me a lead ? Thank you!
at.algebraic topology – Any continuous map is homotopic to one assuming fixed values at finitely many points
Let $X$ and $Y$ be topological spaces. Assume $X$ no dense finite subset and $Y$ is pathconnected. Given $n$ pairs of points $(x_i, y_i)$ where $x_iin X$ and $y_iin Y$ for $1leq ileq n$ and a continuous map $f:Xto Y$ can we find a continuous map $f’:Xto Y$ homotopic to $f$ such that $f(x_i)=y_i$?
ct.category theory – is the inclusion of his 3skeleton in the walking idempotent homotopic cofinal?
To let $ Idem $ be the walking idemotent (1) and let $ Idem ^ {(3)} $ be his 3 skeleton. Note that $ Idem $ has a nondegenerate simplex in every dimension. Lurie has shown (2):
 If $ X $ is a quasi category, then the card $ X to Delta ^ 0 $ has the right lifting ability in terms of absorption $ Idem ^ {(3)} to Idem $;;
That is, although a homotopycoherent idement contains infinitely many coherence data that correspond to the infinite number of nondegenerate simplifications of $ Idem $Nevertheless, it is guaranteed that all this data is available as soon as the first 3 have been found.
The evidence is a bit complicated and I haven't studied it in detail, but what I understand seems to suggest a positive answer to the following
Question: Is the recording $ Idem ^ {(3)} to Idem $ Homotopy cofinal (i.e. it is cofinal in the $ infty $Categorical sense – Depending on how you define this, it may be necessary to replace Joyalfibrant $ Idem ^ {(3)} $ before the question makes sense)?
In fact, I think the answer to the question is "yes" and a proof with "$ bullet $"above, but the evidence I have is somewhat confused. I would prefer direct combinatorial evidence. Unfortunately, I don't believe the recording $ Idem ^ {(3)} $ if anodyne is correct, every proof will have some complications.
One idea would be to use the usual card $ N to Idem ^ {(3)} $, Where $ N $ is the 1skeleton of natural numbers $ mathbb N $ considered as an ordered set. For receiving $ N to mathbb N $ is a categorical equivalence, and that's easy to show $ mathbb N to Idem $ is homotopycofinal according to Quillens Theorem A (since the relevant slice categories are just ordinary 1 categories). By composition, $ N to Idem $ homotopy is cofinal, but this has an effect $ Idem ^ {(3)} $, Through a cofinality cancellation feature to show this $ Idem ^ {(3)} to Idem $ if homotopia is cofinal, it will suffice to show that $ N to Idem ^ {(3)} $ is homotopia cofinal, which sounds simple since then $ N $ and $ Idem ^ {(3)} $ are finite dimensional. But I'm not sure if the relevant slice objects are finitedimensional …
(1) That is, $ Idem $ is the category with an object and a nonidentity morphism $ i $to satisfy the equation $ i ^ 2 = i $, In this question, I identify 1 categories with their nerves, which are quasi categories.
(2) HTT 4.4.5.20 in the current version. This does not appear in the published version of HTT. It appears in older versions of HA as 7.3.5.14, but was moved to HTT when Lurie rewrote the section on Idempotente in HTT.
at.algebraische Topologie – Example: Closed path not homotopic to the path in subset
I'm looking for an example of the following setting:
For an open subset $ U $ from $ M $are both pathconnected, so there is a closed path in $ M $ This is Not homotopic to a closed path in $ U $while $ M / U $ is simply connected.
Explanations:
 $ M / U $ should denote the quotient that contains all the points of $ U $ to a single point, formal $ m_1 sim m_2 $ iff also not $ m_1 = m_2 $ or both $ m_1, m_2 in U $,
 We can set a base point $ u in U $, Then a closed path in $ M $ is a continuous card $ gamma: (0,1) to M $ With $ gamma (0) = gamma (1) = u $,
In the event that it is possible, I would be happy if $ M $ is a manifold.
Thank you in advance.
at.algebraic topology – What is the homotopic fiber from $ X to X_ {hG} $, where is this a spiky homotopy path?
The unsharpened version is simple: the model $ X = EG times X to (EG times X) / G = X ^ {un} _ {hG} $ is a fibrierung with fiber $ G $, But if we go to the top, $ X = EG_ + wedge X to (EG_ + wedge X) / G = X_ {hG} $ there is no more fibrillation: its fiber
changes from $ G $ about nonbase points too $ ast $ above the base point.
Of course, these two cases are still closely related. So maybe there is hope to understand the specific case.
Ask: To let $ G $ Be a finite group (or maybe something more general) and leave $ X $ be pointed $ G $Space.

Is there a good way to understand the homotopy fiber? $ F $ the map $ X to X_ {hG} $ (where they are pointed Homotopy tracks)?

In particular, the homotopic fiber sequence works $ F to X to X_ {hG} $ deloop to a homotopy fiber sequence $ X to X_ {hG} to BF $ (as in the untargeted case)?
ct.category theory – Could we form the homotopy category of a dg category by inverting homotopic invertible morphisms?
To let $ k $ be a field and $ mathcal {C} $ a dg category over $ k $, It is standard to define the homotopy category $ H ^ 0 ( mathcal {C}) $ as the category that contains the same objects as $ mathcal {C} $ but morphisms between two objects $ x $ and $ y $ are defined as
$$
H ^ 0 ( mathcal {C}) (x, y): = Z ^ 0 ( mathcal {C}) (x, y) / B ^ 0 ( mathcal {C}) (x, y).
$$
We could also consider the category $ Z ^ 0 ( mathcal {C}) $ and the class $ W $ of homotopic invertible morphisms in $ Z ^ 0 ( mathcal {C}) $, Slightly more precise,
$$
W = {f in Z ^ 0 ( mathcal {C}) (x, y) text {for some} x text {and} y ~  ~ f text {is invertible in} H ^ 0 ( mathcal {C}) (x, y) }.
$$
We could locate the category $ Z ^ 0 ( mathcal {C}) $ by inverted formally $ W $ and get a category $ W ^ { 1} Z ^ 0 ( mathcal {C}) $, There is a natural bulb $ W ^ { 1} Z ^ 0 ( mathcal {C}) to H ^ 0 ( mathcal {C}) $,
My question is: is $ W ^ { 1} Z ^ 0 ( mathcal {C}) $ equivalent $ H ^ 0 ( mathcal {C}) $?
algebraic topology – union of spaces that are homotopic to wed the sum of the spheres.
Let X and Y be two topological spaces and both are homotopic to a wedgesum of spheres of dimension k.
Question: If X cap Y is homotopic to wed the sum of the spheres of dimension k1, then X cup Y is homotopic to a wedgesum of the spheres of dimension k. Is my statement true?
I believe that it is true, but can not put forward a proper argument.