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21 votes
Accepted

History of the pullback corner notation

In an email to me dated 17 February 1992, Peter Freyd said: I was using a different notation in 1974 in lectures at Montreal. A high school teacher named Butler suggested the right-angle. It was an ...
Paul Taylor's user avatar
  • 8,481
20 votes
Accepted

Why do elementary topoi have pullbacks?

I'll give a counter-example to the claim that having a subobject classifier and being cartesian closed implies the existence of all finite limits. However, this is based on the definition of sub-...
Simon Henry's user avatar
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15 votes

History of the pullback corner notation

If it is of any use, on page 251 of Taylor's "Practical Foundations of Mathematics", shortly after introducing pullbacks, he writes: Pullbacks are often indicated with the right angle symbol, which ...
Christopher Dunkin's user avatar
6 votes
Accepted

Homotopy pullback is right adjoint in the derived category

There is no such functor $\mathcal D/Y\to \mathcal D/X$. It's clear what is meant to be on objects, but it is not well-defined on morphisms. Let $f$ be the inclusion of a point $p$ into a circle $C$. ...
Tom Goodwillie's user avatar
6 votes
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Tensor product of intersections in an abelian rigid monoidal category

Here's a useful lemma to check whether a square in an abelian category is a pullback square: Lemma: A square $A_0\to (A_1,A_2)\to A_3$ with $A_0\to A_1, A_2\to A_3$ being monomorphisms is a pullback ...
Maxime Ramzi's user avatar
  • 15.8k
5 votes
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Intersection of subalgebras in an abelian monoidal category

Yes, such an algebra structure exists. In fact, there is a unique algebra structure on $A \cap B$ such that $j_A : A \cap B \to A$ and $j_B : A \cap B \to B$ are algebra morphisms. To construct the ...
Brian Shin's user avatar
5 votes
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Intersection of Frobenius subalgebra objects

Let $k$ be any field. Consider the commutative $k$-algebra $k[x,y,z]/(x^2,y^2,xz,yz,xy-z^2)$. This is a five dimensional algebra with basis $1$, $x$, $y$, $z$, $u$ where $u=xy=z^2$. It is clearly a ...
Dave Benson's user avatar
  • 16.2k
5 votes
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Homotopy pullback in the category of DG algebras

I was wondering if somebody could tell me the definition of homotopy pullback. Suppose $\def\C{{\cal C}}\C$ is a relative category, i.e., a category equipped with a subcategory, morphisms in which ...
Dmitri Pavlov's user avatar
5 votes

Behavior of divisors under push forward and pull back

Of course, just choose a point on each (divisorial) component of the exceptional locus of $f$, then for each of these points the divisors in $|H|$ passing through form a hyperplane in $|H|$, and any ...
Sasha's user avatar
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4 votes

History of the pullback corner notation

A small comment, that does not fit in its proper place. The department of Algebra in the University of Santiago de Compostela has studied categories since the second half of the sixties under the ...
Leo Alonso's user avatar
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4 votes
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Pullback of localizations

$\newcommand{\Cat}{\mathrm{Cat}_\infty} \newcommand{\Spaces}{\mathrm{Spaces}} \newcommand{\map}{\mathrm{map}} \newcommand{\Fun}{\mathrm{Fun}}$ Any coCartesian fibration which is a localization is a ...
Maxime Ramzi's user avatar
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4 votes
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Properties of codimension under pull back

This is false as stated: taking for $f$ the embedding of a closed subscheme $X\subset Y$, it would mean that $\operatorname{codim}(X\cap Z,X)\leq \operatorname{codim}(Z,Y) $. There are well-known ...
abx's user avatar
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3 votes
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Pullback of monomorphisms between selfdual objects in a tensor category

This is not true in the non-semisimple case. Here is an example from finite group representation theory. Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of ...
Dave Benson's user avatar
  • 16.2k
3 votes

Pull-back a section of a vector bundle

abx is right. This kind of things is built through a partition of the unity. Or if you like better, put a Riemannian structure on $M$. The vector bundle $D^\bot$ orthogonal to $D$ in $TM$ is ...
Gael Meigniez's user avatar
3 votes
Accepted

Does Spec functor sends pushouts of rings into pullbacks of sets?

Let me work directly with schemes (you can of course restrict to affine schemes if necessary). For $S$-schemes $X,Y$ there is a natural map $$|X \times_S Y| \to |X| \times_{|S|} |Y|,$$ which is ...
Martin Brandenburg's user avatar
2 votes

Does "symmetry" of a pullback connection should be obvious?

Here is a proof which is not very elegant, but avoids coordinates. First of all, you define $$\hat T(X,Y)=\nabla_X^{\phi^*TN}d\phi(Y)-\nabla_Y^{\phi^*TN}d\phi(X)-d\phi([X,Y])$$ and observe that this ...
Sebastian's user avatar
  • 6,825
2 votes
Accepted

Are the injections of a coproduct a cover in the canonical pretopology?

This will not be the case in general. A family is a cover in the canonical topology if all its pullbacks are jointly regular epimorphism. So this will for example be the case if coproducts are ...
Simon Henry's user avatar
  • 42.4k
1 vote
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Pullback of Lie algebras

A reference for pullbacks for modules over a ring: Proposition 5.11, p.222 in "An Introduction to Homological Algebra" by Rotman. For Lie $k$-algebras (or just $k$-algebras), you could refer ...
spin's user avatar
  • 2,821
1 vote

Pullback of Lie algebras

This follows from the existence of free Lie algebras. More precisely, the forgetful functor $U: \mathbf{Lie}_k \to \mathbf{Vect}_k$ has a left adjoint $F$ which takes $V$ to the free Lie algebra on $V$...
Joshua Mundinger's user avatar
1 vote

Why trace is more natural than (preferred to) determinant for smooth map $f:M\to N$?

It suffices to understand the special case of a linear map $T:U\to V$ where $U,V$ are Euclidean vector spaces. (Think $U=T_pM$, $V=T_{f(p)}N$, $T=df(p)$.) Suppose first that $n=\dim V\leq \dim U=m$. ...
Liviu Nicolaescu's user avatar
1 vote
Accepted

Injectivity of the cohomology map associated to the pullback of line bundles

The unit of adjunction map $L\rightarrow f_\ast f^\ast L$ is an isomorphism if the fibres of $f$ are connected (I assume this in what follows). Then the Leray spectral sequence $E^{pq}_2=H^p(R^q f_\...
ssx's user avatar
  • 2,808
1 vote
Accepted

Pullback of homogeneous twisted differential operators

This is true. One definition of htdo is analogues to that of equivariant sheaf: a htdo is a tdo $\mathcal D$ equipped with an isomorphism $act^\sharp \mathcal{D} \cong pr_X^\sharp \mathcal{D}$ ...
Qixian Zhao's user avatar

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