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Stream: theory: algebraic topology

Topic: interchange map is mono?

Naso (Feb 27 2022 at 03:29):

In simplicial sets there is an 'interchange' map
$h : (A \times B) * (C \times D) \to (A * C) \times (B * D)$
given by
$h = (\pi_A * \pi_C, \pi_B * \pi_D)$
where $\times$ is the cartesian product and $*$ is the join.
Is it a monomorphism? How can I see that it is or is not? I know the criterion of monomorphism (left-cancellative) but I don't know how to check if it holds here.

Naso (Feb 27 2022 at 05:10):

Nasos Evangelou-Oost said:

In simplicial sets there is an 'interchange' map
$i : (A \times B) * (C \times D) \to (A * C) \times (B * D)$
given by
$i = (\pi_A * \pi_C, \pi_B * \pi_D)$
where $\times$ is the cartesian product and $*$ is the join.
Is it a monomorphism? How can I see that it is or is not? I know the criterion of monomorphism (left-cancellative) but I don't know how to check if it holds here.

Ok, now I'm thinking it is a mono but I'm not completely confident in my reasoning: If $f,g : Q \to (A \times B) * (C \times D)$ and $x \in Q_n$ then since $(X * Y)_n = \coprod_{i+j+1=n} X_i \times Y_j$,
$f(x) := (f(x)_i, f(x)_j) = ((f(x)_i^a,f(x)_i^b), (f(x)_j^c,f(x)_j^d)) \in (A \times B)_i \times (C \times D)_j$
for some $i,j$ with $i + j + 1 = n$.
Now $h(f(x)) = ((f(x)_i^a,f(x)_j^c),(f(x)_i^b), f(x)_j^d)) \in (A * C)_n \times (B * D)_n$.
So if $h(f(x)) = h(g(x))$ then $f(x) = g(x)$ since the dimensions and components must match, so $h$ is a monomorphism?

Naso (Mar 02 2022 at 00:51):

I also asked this on StackExchange (no answer yet) https://math.stackexchange.com/questions/4393365/interchange-map-in-simplicial-sets-is-a-monomorphism

Morgan Rogers (he/him) (Mar 02 2022 at 09:21):

Simplicial sets form a presheaf category. In particular, for a map to be a mono, you only need to check that it's componentwise injective.

Naso (Mar 02 2022 at 12:58):

Morgan Rogers (he/him) said:

Simplicial sets form a presheaf category. In particular, for a map to be a mono, you only need to check that it's componentwise injective.

thanks @Morgan Rogers (he/him) , I do know that, this is what I was trying to check. Sorry if it was not clear

Morgan Rogers (he/him) (Mar 02 2022 at 13:39):

Let's work through it! Explicitly,
$((A \times B) * (C \times D))_n$ is $\coprod_{i+j = n} (A_i \times B_i) \times (C_j \times D_j)$
where I'm taking the shortcut of taking the join of augmenting and then truncating the simplicial sets to simplify the notation. Meanwhile,
$((A * C) \times (B * D))_n = \left(\coprod_{i+j = n} (A_i \times C_j)\right) \times \left(\coprod_{i+j = n} (B_i \times C_j) \right).$
If I'm not mistaken, the map $h$ sends $((a,b),(c,d))$ to $((a,c),(b,d))$, with no identification happening. This is "obviously" injective, since if two simplices in the domain have equal images, their component simplices must be the same.

Naso (Mar 02 2022 at 13:52):

Very clear! Thank you, Morgan :smile: :pray:

Morgan Rogers (he/him) (Mar 02 2022 at 14:43):

I'll copy that across to stackexchange and correct the indices.