Homotopy equivalence re-definition Homotopy equivalence is defined thus: Two spaces $X$ and $Y$ are homotopy equivalent if there are continuous maps $f: X \rightarrow Y$ and $g: Y \rightarrow X$ with $gf \sim \textbf{1}_X$ and $fg \sim \textbf{1}_Y$.
I am rephrasing my question: If we said the image of $f$ can be deformed continuously to $1_Y$ and the same with $g$ deformed to $1_X$.
Would such a definition be even useful?    
 A: If I understood the question correctly, you are asking: if there exist maps $f : X \to Y$ and $g : Y \to X$ such that $Y$ deformation retracts onto the image of $f$, and $X$ deformation retracts onto the image of $g$, then are $X$ and $Y$ homotopy equivalent?
(It's the only reasonable interpretation I can make of the question and the follow-up comments. But to be honest I don't understand why you would expect that to be true: there is absolutely no reason $f$ and $g$ are related in this scenario, and there are immediate counterexamples as you will see. If I got the question wrong, please clarify your question and ask it in precise terms, without any vagueness...)
The answer is no. Let $f : \mathbb{R} \to \mathbb{S}^1$ be the exponential map, and $g : S^1 \to \mathbb{R}$ be a constant map. Then $f$ is surjective, so its image is the whole of $\mathbb{S}^1$ and of course there's a deformation retraction; and $\mathbb{R}$ deformation retracts onto a point (the image of $g$). But $\mathbb{R}$ and $S^1$ are not homotopy equivalent.
