Exercise 6, Section 3.2 of Hoffman’s Linear Algebra 
Let $T$ be a linear transformation from $\Bbb{R}^3$ into $\Bbb{R}^2$, and let $U$ be a linear transformation from $\Bbb{R}^2$ into $\Bbb{R}^3$. Prove that the transformation $U\circ T$ is not invertible. Generalize the theorem.

Potential approach: Assume towards contradiction, $U\circ T$ is invertible. By exercise 4 section 2 of Munkres’ topology, $T$ is injective and $U$ is surjective. We’ll only use $T$ is injective. By this post, $|\Bbb{R}|$ $=|\Bbb{R}^2|$ $=|\Bbb{R}^3|$. Is the following lemma hold: If $f:A\to B$ is injective and $|A|=|B|$, then $f$ is surjective ? Let assume for a moment lemma holds, $T$ is surjective. So $T$ is a bijective linear map. So $\Bbb{R}^3\cong \Bbb{R}^2$. By exercise 6 section 3.3, $\mathrm{dim}(\Bbb{R}^3)$ $= \mathrm{dim}(\Bbb{R}^2)$ $=3=2$. Thus we reach contradiction. Hence $U\circ T$ is not even injective. Can you please verify lemma?

Generalize theorem:

Let $V,W$ be $n,m$-dimensional vector space over field $F$ such that $n\gt m$. If $T\in L(V,W)$ and $U\in L(W,V)$, then $U\circ T$ is neither injective nor surjecctive. In particular, $U\circ T$ is not bijective.

Approach(1): Assume towards contradiction, $U\circ T$ is injective. By exercise 4 section 2 of Munkres’ topology, $T$ is injective. So $N_T=\{0_V\}$. $V$ is finite dimensional vector space. By rank nullity theorem, $n=\mathrm{dim}(V)$ $= \mathrm{dim}(N_T)+ \mathrm{dim}(R_T)$ $= \mathrm{dim}(R_T)$. Since $R_T\leq W$ and $\mathrm{dim}(W)=m$, we have $\mathrm{dim}(R_T)\leq \mathrm{dim}(W)=m$. So $n= \mathrm{dim}(R_T)\leq m$. But $n\gt m$. Thus we reach contradiction. Hence $U\circ T$ is not injective. Assume towards contradiction, $U\circ T$ is surjective. By exercise 4 section 2 of Munkres’ topology, $U$ is surjective. So $R_U=V$. $W$ is finite dimensional vector space. By rank nullity theorem, $m=\mathrm{dim}(W)$ $=\mathrm{dim}(N_U)+ \mathrm{dim}(R_U)$ $= \mathrm{dim}(N_U)+ \mathrm{dim}(V)$ $= \mathrm{dim}(N_U)+n$. So $m= \mathrm{dim}(N_U)+n$. which implies $\mathrm{dim}(N_U)$ $=m-n\lt 0$. Thus we reach contradiction. Hence $U\circ T$ is not surjective. So $U\circ T$ is neither injective nor surjective.
Approach(2): Assume towards contradiction, $U\circ T$ is injective. By exercise 4 section 2 of Munkres’ topology, $T$ is injective. Let $B_V=\{\alpha_1,,…,\alpha_n\}$ be basis of $V$. So $(T(\alpha_1),…,T(\alpha_n))$ is sequence in $W$. If $T(\alpha_i)=T(\alpha_j)$, for some $i,j\in J_n$, then $(T(\alpha_1),…,T(\alpha_n))$ is dependent. If $T(\alpha_i)\neq T(\alpha_j)$, $\forall i\neq j$, then $(T(\alpha_1),…,T(\alpha_n))$ is dependent, since $|\{ T(\alpha_1),…,T(\alpha_n)\}|=n\gt m$. So $\exists (c_1,…,c_n)\in F^n$ such that $c_i\neq 0_F$ for some $i\in J_n$ and $\sum_{i=1}^nc_i\cdot_W T(\alpha_i)=0_W$. Since $\{\alpha_1,…,\alpha_n\}$ is independent, we have $x=\sum_{i=1}^nc_i\cdot_V \alpha_i \neq 0_V$. So $T(x)$ $=T(\sum_{i=1}^nc_i\cdot_V \alpha_i)$ $= \sum_{i=1}^nc_i\cdot_W T(\alpha_i)=0_W$. So $0\neq x\in N_T$. Which implies $T$ is not injective. Thus we reach contradiction. Hence $U\circ T$ is not injective. Assume towards contradiction, $U\circ T$ is surjective. By exercise 4 section 2 of Munkres’ topology, $U$ is surjective. Let $B_W=\{\beta_1,…,\beta_m\}$ be basis of $W$. So $(U(\beta_1),…,U(\beta_m))$ is sequence in $V$. $|(U(\beta_1),…,U(\beta_m))|\leq m\lt n$. By theorem 4 corollary 2 section 2.3, $\mathrm{span}(U(\beta_1),…,U(\beta_m))\neq V$. So $\exists x\in V$ such that $x\notin \mathrm{span}(U(\beta_1),…,U(\beta_m))$. By exercise 8 section 3.1, $\mathrm{span}(U(\beta_1),…,U(\beta_m))=R_U$. Which implies $U$ is not surjective. Thus we reach contradiction. Hence $U\circ T$ is surjective. So $U\circ T$ is neither injective nor surjective.
 A: The lemma is false. Consider the case $A=B=\mathbb{Z}$ and $f:\mathbb{Z}\to \mathbb{Z}$ defined by $f(x)=2x$. Then $f$ is injective but not surjective because $1\notin Im(f)$.
For the exercise problem, just notice that $U$ is surjective will imply that $3=\dim \mathbb{R}^3=\dim \operatorname{Im}(f)=\dim \mathbb{R}^2-\dim \ker(U)=2-\dim\ker(U)$. This is a contradiction. In linear algebra, dimension is a more precise measure of a space  than cardinality.
A: A linear map $T:\Bbb{R}^3\to \Bbb{R}^2$ can't be injective. Since  $\dim(R(T)) \le 2$ , $\dim(\ker T) =3-\dim(R(T)) \ge 1$
Let us choose $0\neq v\in \ker T$ , the
$(U\circ T) (v) =U(T(v))=0$
Hence $\dim (\ker U\circ T) \ge 1$
Since $U\circ T$ is not injective, it can't be invertible.

For the generalization:
A linear map $T:V\to W$  (where $\dim V=n>m=\dim W$ ) can't be injective. Since  $\dim(R(T)) \le m$ , $\dim(\ker T) =n-\dim(R(T)) \ge n-m>0$
Let us choose $0\neq v\in \ker T$ , the
$(U\circ T) (v) =U(T(v))=0$
Hence $\dim (\ker U\circ T) \ge 1$
Since $U\circ T$ is not injective, it can't be bijective and invertible.
(a linear map between two same finite dimensional vector spaces over the same base field is injective iff surjective iff bijective).
