$S:V\to W$ and $T:W\to X$ are two linear maps such that $T\circ S:V\to X$ is injective. Then choose the coorect option

Question

Let $V,W$ and $X$ be three finite dimensional vector space such that $\dim V =\dim X$. Suppose $S:V\to W$ and $T:W\to X$ are two linear maps such that $T\circ S:V\to X$ is injective. Then

Choose the correct option:

1. $S$ and $T$ both are injective
2. $S$ and $T$ both are surjective
3. $S$ is injective, $T$ is surjective
4. $T$ is injective, $S$ is surjective

My attempt:

Since $T\circ S:V\to W$ is injective so $S$ is injective. Let $\dim V=\dim X=n$ and $\dim W=m$ since $S$ is injective then $\dim V<\dim W$ ,then $\dim W<\dim X=\dim V$ which contradict $\dim V<\dim W$.

what will I do?

Answer 1

My attempt Since ToS:V→W is injective so S is injective. Let dim V=dim X=n and dim W=m since S is injective then dimV

Answer 2

You claim that $S$ is injective but you don't claim why (although you are indeed correct). If $S(x)=S(y)$, then $T(S(x))=T(S(y))$ which implies $x=y$ because $T\circ S$ is injective.

Now, are you familiar with the rank-nullity theorem? From this it follows that $T\circ S$ is surjective. Then from there you can use a similar argument as above to show that $T$ is surjective, so that the answer is $(3)$.

Edit: To see that $T\circ S$ is surjective, note that by rank-nullity

$$\dim V=\dim\ker(T\circ S)+\dim\text{im}(T\circ S)=\dim\text{im}(T\circ S)$$

where the last equality is because $\dim\ker(T\circ S)=0$ (because the map is surjective).

Now, we know that $\text{im}(T\circ S)$ is a subspace of $X$, and $\dim\text{im}(T\circ S)=\dim V=\dim X$. Because we are in the finite-dimensional case, this implies $\text{im}(T\circ S)=X$.