$f$ is homotopic to $g$ iff exists a continuous path from $f$ to $g$

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Let $X,Y$ topological spaces. $C(X,Y)$ a set of continuous functions $f:X\rightarrow Y$.

Let $f,g\in C(X,Y)$ homotopic functions, let $H:X\times [0,1]\rightarrow Y$ a homotopy between $f$ and $g$.

We can define $H_f^g:[0,1]\rightarrow C(X,Y)$ with $H_f^g(t)=H(\cdot,t)$.

In $C(X,Y)$ consider the induced strong topology by $\{H_f^g;f\text{ homotopic to }g\}$

Assume that $\pi_x:C(X,Y)\rightarrow Y$ , $\pi_x(f)=f(x)$, is continuous.

Question: $f$ is homotopic to $g$ iff exists $\varphi:[0,1]\rightarrow C(X,Y)$ continuous path from $f$ to $g$.

$(\Rightarrow)$It's clear why $C(X,Y)$ has an induced strong topology .

$(\Leftarrow)$ (problematic part)

I define $H(x,t)=\pi_x(\varphi(t))=e\circ (\operatorname{Id}_X\times \varphi)(x,t)$, where

$\operatorname{Id}_X\times \varphi:X\times [0,1]\rightarrow X\times C(X,Y)$ is continuous.

$e:X\times C(X,Y)\rightarrow Y$, $e(x,f)=f(x)$. I can't prove that $e$ is continuous.

Without adding more assumptions, is it possible to demonstrate continuity or is there a counterexample?