1
$\begingroup$

If $X$ is a simply connected space and $\alpha\in\pi_n(X)$ and $Y=X\cup_{\alpha} e^{n+1}$. and $(\wedge V,d)$ be the minimal model for $X$,then $(\wedge V\oplus \mathbb{Q} u,d)$ is a commutative model for $Y$, which we denote by $M_\alpha$. If $f\colon X\rightarrow X_1$ and $Y=X\cup_{\alpha}e^{n+1}$ and $Y_1=X_1\cup_{f^*(\alpha)}e^{n+1}$,then we have a canonical map $g\colon X_1\rightarrow Y_1$ and the following commutative diagram (1)

\begin{array}{lll} X & \rightarrow &Y\\ f\downarrow && \downarrow g\\ X_1 & \rightarrow & Y_1 \end{array}

Now,if $(\wedge V_1,d)$ is the minimal model for $X_1$ and so $(\wedge V_1\oplus \mathbb{Q}u,d)$ is a commutative model for $Y_1$ and if $\phi\colon (\wedge V_1,d)\rightarrow (\wedge V,d)$ be the Sullivan representative for $f$,then we have induced map $\psi$ from $(\wedge V_1\oplus\mathbb{Q}u)$ to $(\wedge V\oplus\mathbb{Q}u,d)$which is $\phi$ on $(\wedge V_1)$ and identity on $u$.

Then we have the following diagram (2)

\begin{array}{lll} (\wedge V_1\oplus\mathbb{Q} u,d) & \rightarrow &(\wedge V_1,d)\\ \psi\downarrow & & \downarrow \phi\\ (\wedge V\oplus\mathbb{Q}u,d) & \rightarrow &(\wedge V,d)\\ \end{array}

Now,my questions are

a. Is $\psi$ a commuatative model for $g\colon Y\rightarrow Y_1$?

b. Is (2) a commutative model for (1)?

$\endgroup$

1 Answer 1

1
$\begingroup$

In rational homotopy theory models of homotopy push-outs are homotopy pull-backs in the category of CDGAs. This fact follows from the Quillen adjunction between model categories of spaces and CDGAs. In fact this is just a Mayer-Vietoris type argument.

(1) is a homotopy push-out: $X\rightarrow Y$ is a cofibration and (1) is a push-out. To get an algebraic model of (1) you should build a surjective model of the map $f:X\rightarrow X_1$ and then take the pull-back along the model of the map $X\rightarrow Y$ and you will get a model of (1). And that's what you have done.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .