# Contraction of a family of loops simultaneously

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 $$t\in [0,1]$$?

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.