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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.