Given a linear map $T:V\rightarrow W$ it is true that:
1A)If $\dim(V) > \dim (W)$ then $T$ is Not injective to which the contrapositive is
1B)If $T$ is injective then $\dim (V) \leq \dim (W)$
and
2A)If $\dim(V) < \dim(W)$ then $T$ is Not surjective to which the contrapositive is
2B)If $T$ is surjective then $\dim (V) \geq \dim (W)$
It is also true that
3A)if $\dim(V) = \dim (W)$ then ($T$ is injective iff and only if $T$ is surjective) of which the contrapositive is
3B)($T$ is Not injective and $T$ is surjective) or ($T$ is injective and $T$ is not surjective) then $\dim (V) \not= \dim (W)$
Now 3B tells us that ($T$ is Not injective and $T$ is surjective) or ($T$ is injective and $T$ is not surjective) only occurs when $\dim (V) \not= \dim (W)$ , which can be expressed as when ($\dim(V) > \dim (W)$) or ($\dim(V) < \dim(W)$).
So does combining 3B with 2B allow us to conclude that:
If $\dim(V) > \dim (W)$ then $T$ is Not injective but is surjective
and combining 3B with 1B
If $\dim(V) < \dim(W)$ then $T$ is Not surjective but is injective
No. If $\dim V>\dim W$, then $T$ is not injective, but it doesn't have to be surjective. Just assume that $\dim V>0$ and let $T$ be the null function.
And if $\dim V<\dim W$, then $T$ is not surjective, but it doesn't have to be injective. Again, take the null function.