When one can extend $X\times \{0,1\}\cup A\times[0,1]\rightarrow Y$ to full homotopy? I'm wondering if there is some general criteria/theorems to answer the following question:
Question: Let $X$ and $Y$ be topological spaces (even metric spaces if that is necessary). Let $A\subseteq X$ be a closed subspace of $X$. If we have a continuous map
$$H:X\times \{0,1\}\cup A\times[0,1]\rightarrow Y$$
Then, when can one extend this $H$ to full homotopy?
$$\overline{H}:X\times[0,1]\rightarrow Y$$
-The above question arose while I was considering the special case where $X=Y$ and $A$ is a singleton set in $X$ to show some homotopy equivalence.
-This is very similar situation with homotopy extension property, but more restrictive since the values are already given when $t=1$. Do we still have some useful criteria/theorems to get a full homotopy in this situation?
 A: This is a special case of The Extension Problem: for $X$ and $Y$ CW complexes and $f\colon A\to Y$ a continuous function defined on a subcomplex $A\subset X$, we want to solve the problem of "When can $f$ be continuously extended to all of $X$?" The general solution is something called Obstruction Theory 
which gives us cohomology classes 
$$ \mathfrak{o}_r(f) \in H^{r+1}(X,A;\pi_r(Y)) \cong \tilde{H}^{r+1}(X/A;\pi_r(Y)) $$
for each $r$, so that $f$ can be extended iff these classes all vanish.
The modern formulation is in Hatcher, and is in terms of a Moore-Postnikov system for $f$ and is a little advanced. The classical (more technical to set up but only involves cellular cohomology) version has a good presentation in Steenrod: the rough idea is that, supposing that $f$ has an extention to the $r$-skeleton $f_r\colon X^{(r)} \cup A \to Y$, it can be extended over an $(r+1)$-cell $D$ in $X$ iff $D\subset A$ or $f_r(\partial D)$ is null-homotopic in $Y$, i.e. trivial in the group $\pi_r(Y)$ (you need to take care with basepoints). The collection of these $[f_r(\partial D)]$ for each $(r+1)$-cell $D$ in $X$ miraculously combines into a relative cohomology class in degree $r+1$.
Now if we have two functions $f_0, f_1\colon X \to Y$ and a homotopy defined on $A$, these piece together into a function
$$ F\colon (X\times\{0,1\}) \cup (A\times I) \to Y$$
which we want to extend to $X\times I$. If $A$ contains the basepoint of $X$ then the quotient by this subcomplex is actually the reduced suspension
$$ (X\times I)/(X\times\{0,1\} \cup A\times I) \cong \Sigma(X/A) $$
therefore by the suspension isomorphism our obstructions live in the groups
$$ \mathfrak{o}_r(F) \in  \tilde{H}^r(X/A;\pi_r(Y)).$$

These obstructions are in practice difficult to compute, and this theory works best when the obstruction groups vanish. For example, this theory is used to show that an Eilenberg-MacLane space $K(G,n)$ (a space with the property that $\pi_nX \cong G$ and $\pi_i X= 0$ otherwise) is unique up to homotopy equivalence.
