Can the loops in the definition of the fundamental group be considered injective?

Question

Let $\mathrm{С}$ be some class of topological spaces that includes at least all subspaces of $\mathbb{R}^n $. Further we are in the category $\mathrm{С}_{*}$ (the category of point spaces; all continuous maps and homotopies preserve the chosen base points). We define $π'_1(X) $ as the set of homotopy classes of products of injective loops in $X$ with the operation $[a] \cdot [b] = [a \cdot b]$. We say that $\mathrm{D} \subset \mathrm{C}$ is essentially wide if for every $X \in \mathrm{C}$ there exists $Y \in \mathrm{D}$ such that $X$ is homotopy equivalent to $Y$.

Is there a $ D $ essentially wide subclass of $ C $ such that, for $ X \in D $

1. The operation is well defined, i.e. there always exist an injective loop homotopic to $a \cdot b$
2. The natural embedding of $π'_1 (X) \to π_1 (X)$ is an isomorphism.

We define $π''_1 (X)$ similarly, but with homotopies in the class of injective loops.

Is there a $ D $ essentially wide subclass of $ C $ such that, for $X \in D$

3. The operation is well defined, i.e. among the classes of injective homotopy there exists and is uniquely a class of loops freely homotopic $a \cdot b$
4. The natural embedding $π''_ 1(X) \to π_1 (X)$ is an isomorphism.

Answer 1

In the Griffiths Twin Cone (or double cone over the shrinking wedge of circles), $G\subseteq \mathbb{R}^3$, all injective loops are null-homotopic yet $\pi_1(G)$ is uncountable. Hence, injective loops don't generate any of the fundamental group. However, as @WillSawin mentions in the comments, it is possible that you can pass to $G\times C$ where $C$ is contractible. This should mean that your first question has a positive answer at least in the case where $C$ is the set of subsets of $\mathbb{R}^n$.

Specifically, given a space $X$ in your class of spaces, we can consider $Y=X\times [0,1]^2$ which is homotopy equivalent to $X$ by the projection and seems to be in the class you are interested in. Given a loop $\alpha:(S^1,\ast)\to (X,x_0)$, we can define paths $\beta_1,\beta_2:[0,1]\to [0,1]^2$ so that $\beta_1(t)=(t,0)$ and $\beta_2:[0,1]\to [0,1]^2$ is some other arc from $(0,0)$ to $(1,0)$ that only meets $[0,1]\times\{0\}$ at its endpoints. Now consider the path $(\alpha(t),\beta_1(t))$ in $Y$ and follow it by the path $(\alpha(\ast),\beta_{2}(1-t))$. This concatenation defines an injective loop in $Y$ whose projection to $X$ is homotopic to $\alpha$. Hence, the injective loops in $Y$ generate the entire fundamental group of $Y$.

Answer 2

For the new version of the question (where you allow to replace the space by a homotopy equivalent one) the answer is now "yes": just replace every $X$ by $\lvert\operatorname{Sing}(X)\rvert$. This is homotopy equivalent at least for $X$ a CW complex, and since $\operatorname{Sing}(X)$ is a Kan complex, every element in $\pi_1$ is represented by an embedded loop.