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I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1. A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1. A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1. A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1. A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Clarification: A pullback in a certain category is defined as a space satisfying a universal property: a space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$, there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

Remarks:

1. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

2. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

3. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1. A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2. I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3. In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4. In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).