Contraction of a family of loops simultaneously

Question

Let $LX$ denote the free loop space on $X$. We have an evaluation map $ev\colon LX\to X$ and we have an inclusion $X\hookrightarrow LX$ (where $x\in X$ is mapped to the constant loop at $x$).

Suppose that $A\subset X$ is a cofibration, where $X$ is simply connected (I wouldn't mind to assume that $A$ is also simply connected if that would simplify anything). Let $\varphi\colon A\to LX$ be a map that makes the following diagram commutes:

In other words: $\varphi$ sends $a\in A$ to a loop in $X$ that is based in $a$.

My question: Is it possible to contract all loops simultaneously but in a "basepoint preserving" way (see below)? I.e., is is possible to find a homotopy $\psi_t\colon A\to LX$ where $\psi_0=\varphi$ and $\psi_1$ is the inclusion $A\hookrightarrow LX$, where

commutes for every $t\in [0,1]$?

Answer 1

Surely not.

Let $S^1 \to LS^2$ be adjoint to the map $c: S^1 \times S^1 \to S^2$ which collapses $S^1\vee S^1$ to a point. The latter has degree one.

Let $p\in S^1$ be any point but the basepoint. Take $A\to X$ to be the composite map $$ S^1 \times p \subset S^1\times S^1 \overset{c}\to S^2 $$ (this is an inclusion).

If a homotopy of the kind you are requesting existed, then we could conclude, by taking adjoints, that $c: S^1 \times S^1 \to S^2$ is homotopic to a map that factors through $S^1 \times p$. This gives a contradiction, since such a map has degree zero.