I'd like to prove the following, to well define homotopy classes and essentialy work with the most useful space, (i.e being able to switch between $(I,\partial I)$ and $\mathbb{S}^1$):
Lemma : A continuos map $f: (X,A) \longmapsto (Y \star)$ into a pointed space induces a pointed map $\bar{f} : X/A \longmapsto Y$. This assignment $f \to \bar{f}$ induces a bijection $[((X,A)),((Y \star))]\simeq[X/A,Y]^0$.
I was able to prove the first part of the Lemma but I don't understand the symbol $\simeq$. Here denotes an homeomorphism or just bijection?
The hint to prove the Lemma is the following Proposition which I don't know how to use in order to prove the Lemma :
Proposition : Let $p: X \longmapsto Y$ be a quotient map. Suppose $H_t:Y \longmapsto Z$ is a family of set maps such that $H_t \circ p$ is a homotopy. Then $H_t$ is an homotopy.
Any help or reference on the proof of both statemente would be appreciated.
It denotes bijection.
Let $p : X \to X/A$ denote the quotient map. Define $$p^* : C(X/A,*) \to C((X,A),(Y,*)), p^*(f) = f \circ p .$$ Here $C$ denotes the set of continuous maps of pairs. Note that $C((U,u_0), (V,v_0))$ is nothing else than the set of basepoint-preserving maps.
Clearly $p^*$ is a bijection; this is due to the fact that that $p$ is a quotient map. Simply note that each $\phi : (X,A) \to (Y,*)$ has the property $\phi(A) = *$ so that it induces a unique $\bar \phi : X/A \to Y$ such that $\bar \phi \circ p = \phi$. This map has the properyt $\phi(*) = *$.
Clearly $p^*$ maps homotopic maps to homotopic maps, i.e. induces $$p^* : [(X/A,*),(Y,*)] \to [(X,A), (Y,*)], p^*([f]) = [p^*(f)] .$$ This map is surjective by definition. Now let $p^*([f_0]) = p^*([f_1])$. This means that there exists a homotopy of pairs $H : (X,A) \times I = (X \times I, A \times I) \to (Y,*)$ such that $H_0 = f_0, H_1 = f_1$. It has the property $H(A \times I) = *$. It is well-known that $p \times id_I : X \times I \to (X/A)\times I$ is a quotient map. Hence $H$ induces a unique $\bar H : (X/A)\times I \to Y$ such that $\bar H \circ(p \times id_I) = H$. Clearly $\bar H$ is a homotopy from $f_0$ to $f_1$. That is, $[f_0] = [f_1]$ which proves that $p^* : [(X/A,*),(Y,*)] \to [(X,A), (Y,*)]$ is a bijection.