Is it possible for a closed manifold to deformation retract onto a proper subset of itself?

Let $M$ be a closed (compact, without boundary) topological manifold. Is it possible for there to exist a subset $A$ of $M$ such that $M$ deformation retracts onto $A$?

I'm not 100% sure of my answer. I'll assume $M$ to be connected; $n$ stands for its dimension. Suppose $M$ deformation retracts onto a proper subet $A$.

Because $M$ deformation retracts onto $A$, the long exact homology sequence (LEHS) of the triple $\emptyset\subset A\subset M$ with coefficients in $G=\Bbb Z$ or $G=\Bbb Z/2\Bbb Z$ (depending on wether $M$ is orientable or not) tells us $$H_*(M,A)\equiv 0.$$

Using the above and the LEHS for $A\subset M\setminus\text{pt}\subset M$, there are isomorphisms $$H_{*-1}(M\setminus\text{pt},A)\simeq H_*(M,M\setminus\text{pt})\simeq H_*(\Bbb R^n,\Bbb R^n\setminus 0).$$ The first one is the connecting homomorphism, the second one arises from excision. Therefore $H_*(M\setminus\text{pt},A)=0$ for $*=n,n+1$ Then, the LEHS of the triplet $\emptyset\subset A\subset M\setminus \text{pt}$ in degree $n$ tell us that

$$H_n(A)\simeq H_n(M\setminus \text{pt})$$

Since $M$ is closed connected, its top homology is isomorphic to $G$, and so is that of $A$ since they are homotopy equivalent: $$H_n(M)\simeq H_n(A)\simeq G.$$ Also, since $M\setminus\text{pt}$ is a non compact connected manifold, its top homology is $0$: $$H_n(M\setminus\text{pt})=0.$$

This contradicts the isomorphism $H_n(A)\simeq H_n(M\setminus \text{pt})$. Thus there are no deformation retractions of $M$ onto a proper subset.

By a corollary of the generalized Poincaré duality, referred as Poincaré Alexander Lefschetz duality in the book Geometry and Topology of Bredon, you can derive the following tool :

Corollary 8.5 (Bredon - G. & T.) If $L$ is a proper compact subset of an orientable connected $n$-manifold $M$, then $\check H^n(L; G) = 0$ for any coefficient group $G$.

The same is true with $\mathbb Z/2$ coefficients if $M$ is not orientable. If $M$ retracts on a proper subspace $L$, then $L$ is closed.

But for any proper closed subset $L$ of $M$, by this property from Bredon, we have $\check H^n(L;\mathbb Z/2)=0$. This implies that $L$ can't be a deform retract of $M$, otherwise $L$ would have a non vanishing top homology, as $M$ does.